A n h sang vifa c6 t i n h chat song n h u giao thoa, n h i l u xa vifa c6 t i n h chat hat n h u hien tupng quang dien ...
Van de 3: H I E N T U O N G Q U A N G D I E N T R O N G - Q U A N G D I E N TRd - P I N Q U A N G D I E N 1. Hi§n tUdng quang di^n b§n trong
a) Dinh nghia: H i e n tupng tao thanh eac electron dan va lo trong trong ban dan, do tac dung cua anh sang thich hop X < X„ hay £> A .
b) Hien tugng quang ddn: H i e n tUpng giam dien t r d suat cua chat b a n d a n lam tang dp d i n dien cija ban dSn.
2. Quang di$n tr6: Dupe che tao dua tren hieu Ung quang dien trong. Do la mot tam ban dan c6 gia t r i dien t r d thay doi k h i cuorng dp chum anh sang chieu vko n6 thay doi.
3. Pin quang di^n: Nguon dien trong do quang nSng dupe bien doi true tiep t h a n h dien nSng. Hoat dpng cua p i n dUa t r e n hien tUpng quang dien ben ciia mot so chat ban d i n . Suat dien dpng ciia p i n thucfng c6 gia t r i tCr 0,5 V den 0,8 V. Pin quang dien (pin M a t Trcfi) la nguon d i ^ n pho bien.
Van de 4 : H I E N T U O N G Q U A N G - P H A T Q U A N G 1. Hi§n tUdng quang - phSt quang
a) Su phat quang
- Co mot so chat sau k h i hap thu anh sang c6 budc song nay, de phat ra anh sang khac c6 budc song Idn hon, gpi la hien tUpng quang - phat quang.
- M o i chat phat quang c6 mot quang phd dac trung cho no.
- Sau k h i ngifng kich thich, sU phat quang cua mot so' chat con tiep tuc keo dai them mot thdi gian nao do, roi m d i ngimg h i n . Khoang thdi gian tif luc ngirng kich thich cho den liic ngifng phat quang gpi la thdi gian phat quang.
b) Huynh quang va Idn quang
+ Su huynh quang la sU phat quang c6 thdi gian phat quang ng^n (dudi 10~'*s). Nghia la anh sang phat quang hau n h u t ^ t ngay sau k h i t ^ t anh sang kich thich. No thudng xay ra vdi chat long va chat k h i .
+ Sir Ian quang la su phat quang c6 thcri gian phat quang dai (tii 10"*s trd len); no thucrng xay ra vdi chat ran. Cac chat ran phat quang loai nay gpi la chat Ikn quang.
'2. DSc dilm cua sa phSt quang
A n h sang phat quang c6 budc song X' dai hcfn bUdc song cua anh sang kich thich X. iX' > X.)
3. Ong dung cija hi^n tUdng phSt quang
SLT dung trong cac den ong de thap sang, trong cac mkn hinh cua dao dpng k i dien tu, tivi, may tinh, s\X dung scfn phat quang quet tren cac bien bao giao thong.
Van de 5: M A U N G U Y E N T L / B O - Q U A N G P H O V A C H C U A N G U Y E N TLT H Y D R O
1. M i u nguyen tCl Bo | a) Cdc tien de cua Bo
* T i e n de v e t r a n g t h a i dUng
N g u y e n tCr c h i t o n t a i t r o n g m o t so t r a n g t h a i c6 nSng li/cmg xac d i n h E n , goi la cac t r a n g t h a i ditog. K h i cf t r a n g t h a i diJng, nguyen ttr k h o n g btifc xa.
T r o n g cdc t r a n g t h a i dCrng cua n g u y e n tuf, e l e c t r o n c h u y e n d o n g q u a n h h a t n h a n t r e n n h u f n g q u y d a o c6 b a n k i n h h o a n t o a n xac d i n h g o i l a quy dao diTng.
B a n k i n h q u y d a o d i t o g cua e l e c t r o n : rn - n^r,), v d i ( n = 1 , 2, 3...) va ro = 5 , 3 . 1 0 ~ " m , g o i l a b a n k i n h B o . D o c h i n h l a b a n k i n h q u y dao dCrng ciia e l e c t r o n , iJng v d i t r a n g t h a i cO b a n .
B i n h t h u ' d n g , n g u y e n tuf a t r a n g t h a i dCrng c6 n S n g l i i d n g t h a p n h a t g o i l a t r a n g t h a i ccf b a n . K h i h a p t h u n S n g lUcJng t h i n g u y e n tuT c h u y e n l e n t r a n g t h a i d i i n g c6 n S n g liTcfng cao h o n , g o i l a t r a n g t h d i k i c h t h i c h . T h d i g i a n n g u y e n tuf d t r a n g t h a i k i c h t h i c h r a t n g ^ n ( c h i cor 10"**s).
Sau do n g u y e n tuT c h u y e n ve t r a n g t h a i dCrng c6 n S n g lircfng t h a p h o n v a cuoi c u n g ve t r a n g t h a i ccf b a n .
* T i e n de ve sU bufc x a v a h a p t h u n S n g l i f o n g ciia n g u y e n tuf
K h i n g u y e n tuf c h u y e n t\i t r a n g t h a i d i t o g c6 n a n g l i i g n g £ „ s a n g t r a n g t h a i diTng c6 n S n g l i r o n g E ^ n h o h o n t h i n g u y e n tiif p h d t r a m p t p h o t o n CO n a n g luomg: e = hfn„ = E „ - E ^ .
NgiTcfc l a i , n e u n g u y e n tuf d a n g cf t r a n g t h a i d i t o g c6 n a n g liicfng E m m a h a p t h u duoc m o t p h o t o n c6 n a n g l i i o n g h f d i i n g b k n g h i e u - E n , t h i no c h u y e n s a n g t r a n g t h a i d i t o g c6 n a n g liicJng E n l<Jn h o n .
Sir c h u y e n tiT t r a n g t h a i d i i n g E m s a n g t r a n g t h a i d i t o g E n l i n g v d i sii n h a y ciia e l e c t r o n tCr q u y dao d i t o g c6 b a n k i n h r ^ s a n g quy d a o d i t o g CO b a n k i n h Tn v a ngucfc l a i .
2. Quang phd vach p h ^ t xa h^p t h u cua nguygn tCf hydrfi
+ Q u a n g pho p h a t x a : K h i e l e c t r o n c h u y e n tiT miifc n a n g l i r o n g cao (Ecao) xuo'ng miJc n a n g l i f o n g t h a p hcfn (Eti,a,>) t h i no p h a t r a mqt p h o t o n c6 n a n g l i i o n g h o a n t o a n xdc d i n h : hf = Ecao - E thap-
c
M 6 i p h o t o n c6 t a n so f ufng v d i m o t dcfn s^c c6 biidc s 6 n g X = —, tao thanh m o t v a c h q u a n g p h o c6 m o t m a u n h a t d i n h . T a p hcfp cdc v a c h q u a n g p h o t r e n diicfc g o i l a q u a n g p h o v a c h cua n g u y e n tuf H y d r o .
+ Q u a n g pho h a p thu: NgUOc l a i , neu m o t nguyen t i r h y d r o d a n g d mbt miJfc nSng liTcfng E^hap nao do m a nhm t r o n g m o t c h i i m a n h sang t r a n g , t r o n g do c6 tat ca cac p h o t o n c6 n d n g lucfng t i i Icfn den n h o khac n h a u , t h i lap tufc nguyfe"
tif do se h a p t h u ngay m o t p h o t o n c6 n d n g li/ong p h u h o p z = Ecao -EthSp chuyen l e n miifc nSng liTong Ecao- N h u vay da c6 m o t dcfn s^c b i h a p t h u I k m cho t r e n n e n quang pho h e n tuc xuat hign m p t vach t o i .
3. Quang phd v?ch cue nguySn tCf H y d r a
00 ,
6 5
t i m c h a r
l a m O
N
do M •
P A S E N
A n h s a n g h o n g n g o a i
B A N M E
m o t p h a n a n h s a n g k h a k i e n , m o t p h a n a n h s a n g tuf n g o a i
K L Y M A N K
K A n h s a n g tuf n g o a i
I Van de 6: H A P T H y V A P H A N X A L Q C L l / A A N H S A N G
ằ - M A U S A C C A C V A T
HM^P t h u 6nh s^ng
r H a p t h u a n h s a n g l a h i e n tiigtng circfng dp c h u m s a n g b i m o i t r u d n g v a t c h a t l a m g i a m k h i a n h s a n g t r u y e n qua m o i t r U d n g .
' a) Dinh luat ve sU hap thu anh sang
; Ci/cfng do I c u a c h u m s a n g d o n s^c k h i t r u y e n q u a m o i trircfng h a p t h u , g i a m t h e o d i n h l u a t h a m m u cua dp d a i cua d i i d n g d i d.
I = lo.e--^'
V d i : IQ: C u d n g dp cua c h u m s a n g t d i m o i tru'cfng.
a : H e so h a p t h u ciia m o i triicfng.
b) Hap thu Igc liia
- M o i c h a t d e u h a p t h u c6 c h p n Ipc d n h s d n g .
- N h i j f n g c h a t h a u n h i i k h o n g h a p t h u a n h s d n g t r o n g m i e n n a o ciia q u a n g p h o diipc g o i l a g a n t r o n g suot vdri m i e n q u a n g p h o do.
- NhCifng v a t k h o n g h a p t h u a n h s a n g t r o n g m i e n n h i n t h a y c i i a q u a n g p h o diipc g o i l a v a t t r o n g s u o t k h o n g m a u .
- N h S n g v a t h a p t h u t a t ca m o i d n h s a n g n h i n t h a y se c6 mku d e n . - V a t t r o n g suot c6 m a u l a n h i J n g v a t ha'p t h u loc l u a a n h s d n g t r o n g
m i e n n h i n t h a y .
2. Phcin xa (hoac tan xa) ioc iua. M^u s5c Ceic v$t
a) P/idn xg (hoac tan xg) Igc lUa.
Mot so vat CO kha nSng phan xa (hoSc t d n xa) anh sang m a n h hoac yeu khac nhau la do phu thuoc vao bifdc song anh sang t d i duoc goi la sir phan xa (hoac t d n xa) loc lUa.
b) Mau sac cdc vgt.
Cac vat the khac nhau c6 mau sdc khac nhau la do cac vat nay dugc cau tao iii nhCirng vat lieu khac nhau. Ngoai ra, mau s^c ciia cac vat con phu thuoc vao mau s^c cua anh sang roi vao no.
Van 6e 7: S O L U d C V E L A Z E
1. Laze: Mot nguon sang phat ra mot chum sang cifdng do Idn difa t r e n viec ufng dung hien tuong phat xa cam ufng.
• Dac diem: Tia Laze c6 t i n h dan sSc, t i n h d i n h hiTdng, t i n h ket hop cao va cUdng do Idn.
Tuy theo vat lieu phat xa, ngifcfi ta da tao ra laze rSn, laze k h i va laze ban dan. Laze rubi (hong ngoc) thuoc loai laze rSn.
• Lfng dung:
- Tia laze c6 ifu the dSc biet trong thong t i n lien lac v6 tuyen (nhu truyen thong thong t i n b^ng cap quang, v6 tuyen d i n h v i , dieu khien con tau vu t r u , ...)
- Tia laze diroc dung nhir dao mo trong phau thuat m^t, d§' chiJa mot so benh ngoai da (nhd tac dung nhiet), ...
- Tia laze duoc dung trong cac dau doc dia CD, but chi bang, chi ban do dung trong cac t h i nghiem quang hoc d triTdng pho thong, ...
- Ngoai ra tia laze con difcfc dung de khoan, cdt, t o i , ... chinh xac cac vat lieu trong cong nghiep.
2. SU ph^t xa c^m Qng: Neu mot nguyen tur dang d trong trang t h a i kich thich, sSn sang phdt ra mot photon c6 nSng liTOng e = hf, bat gap mot photon c6 nang lifong e' dung bhng hf bay \\i6t qua no, t h i lap i<ic nguyen ti]f nay cung phat ra photon e. Photon s c6 ciing nang luong va bay cung phiJcfng vdfi photon e'. Ngoai ra song dien tCr lirng vdi photon e hoan toan cCing pha va dao dong trong mot mat phSng song song vdi mat phing dao dong ciia song di^n tCf ijfng vdi photon e'.
Nhir vay, neu c6 mpt photon ban dau bay qua mot loat cdc nguyen tijf dang d trong trang t h a i kich thich t h i so photon se tang len theo cap so n h a n ^
TRAC NGHIfM LI THUYET
C a u 1. H i e n tuang nao diTdi day \k hien quang dien?
A. Electron bufc ra khoi k i m loai k h i c6 anh sang thich hop chieu v^o k i m loai B. Electron bi bat ra k h o i k i m loai k h i k i m loai bi nung nong.
C. Electron bi bat ra k h o i mpt nguyen tur k h i nguyen tiif n a y iuoiii4 t a c v d i nguyen t\i khac
D. Electron bi bat ra khoi k i m loai k h i c6 ion am hoSc ion duong dap vao kim loai do
C a u 2. K i m loai Kali (K) c6 gidi han quang dien la 0,55pm. H i e n tuong quang dien khong xay ra k h i chieu vao k i m loai do buTc xa nkra trong vung
|. A. Hong ngoai B. Tia Rcmghen C. Anh sdng mau t i m D. Tilr ngoai
, C a u 3. Lan Itrgt chieu hai burc xa c6 budc song "k^ = 0,66pm va X2 = 0,35pm vao I mot t a m k i m loai c6 gidi han quang dien ^0 = 0,35pm. BiJc xa nao gay ra
* hien tifong quang dien
A. Ca hai biirc xa B. Chi c6 buTc xa ki C. Chi CO bufc xa
D. Khong CO biJc xa nao trong 2 biJc xa t r e n
a u 4. Chieu t d i be mat ciia mot k i m loai bufc xa c6 bi/dc song k nSng liTcfng photon la s, gidi han quang dien cua k i m loai do la ^0 va cong thoat cua cdc electron ra khoi k i m loai la A. Biet hhng so Plang la h , van toe dnh sdng V trong chan khong la c. De c6 hien tiTOng quang dien xay ra t h i
•!A.k>ko B. E < A C.'e >A D. ^ < he
^0
C a u 5. Gidi han quang dien cua mot k i m loai xesi la 0,50pm. H i e n tMng quang dien se xay ra k h i chieu vao k i m loai do
A. Tia hong ngoai B. Bufc xa mku do c6 budc song la 0^ '.''.j.ii.i C. Tia tijf ngoai ' D. Bufc xa mau cam c6 bifdc song 0,59,.jni C a u 6. Goi birdc song k„ la gidi han quang dien ciia mot k i m loai, k la bifdc
song anh sang kich thich chieu vao k i m loai do, de hi?n tifgrng quang di$n xay ra t h i
' A. Chi can dieu k i ^ n < ^0.
B. Chi can dieu k i | n k > ko
C. Phai CO ca hai dieu kien k>kovk cKdng dp anh sdng kich thich phai Idn D. Phai CO ca hai dieu kien k > ko va cudng dp anh sang kich thich phai nhd 2au 7. Trong t h i nghiem vdi te bao quang dien, phat bieu nao dudi day dung?
A. Vdi cac k i m loai khac nhau daac diing lam Catot deu c6 ciing mot gidi han quang dien xdc d i n h
B. Khi CO hien tiTcfng quang dien, cudng do dong quang di#n bao hoa t i 1$
thuan vdi cudng do cua chiim sang kich thich
C. Lfng vdi moi k i m loai diing lam Catot, gia t r i ciia hi$u dien the h a m khong phu thuoc vao ban chat ciia k i m loai lam Catot
D. Cong thoat ciia electron k h o i mpt mat k i m loai duac dung lam Catot phu thuoc vao budc sdng dnh sang kich thich
C&u 8: Hi#n ti/Ong quang dan 1^ hi$n tiTcfng khi c6 dnh anh thich hap chie'u vao chat ban dan thi
A. dien trd ciia chat bdn dan giam manh B. so lirpng cac hat mang dien giam
C. sinh ra cdc hat mang dien bat ra khoi chat bdn dSn D. giam do dan dien cho chat ban dan
Cau 9. Khi chieu Ian lucft hai bufc xa c6 h\X6c song la Xi, X2 (vdi X^ < X2) v^u mot qua cau kim loai dat c6 lap t h i deu xay ra hien tucfng quang dien vaj di§n the cue dai eua qua eau Ian lupt la Vi, V2. Neu chieu dong thori hai bilc xa tren vao qua cau nay thi dien the cue dai eua no la
A'. V2 B. (Vi = V2) C. v'l D. I V i - V21
Cau 10. Mot chum anh sang don sac tac dung len be mat mot kim loai va lam biJe cdc electron ra khoi kim loai nay. Neu tang cUcfng dp ehum sang do len gap hai Ian
A. Dong nSng ban dau cue dai ciia eleetron quang dien tang hai Ian B. Dong nang ban dau eUc dai cua electron quang dien tang bon Ian C. Cong thoat eua electron giam hai Ian
D. So li/png eleetron thoat ra khoi be mat kim loai do trong mSi giay tang len 2 i a n
Cau 11. Trong t h i nghiem vdi te bao quang dien, khi chieu chum sang kich thich vao Catot thi eo hien tupng quang dien xay ra. De triet tieu dong quang dien, ngUbi ta dat vao giOa Anot va Catot mot hieu dien the gpi la hieu dien the ham. Hieu dien the ham nay c6 dp Idn
A. Phu thupc vao nSng lifpng cua chum sang kich thich B. Lam tSng to'c electron quang dien di ve Anot
C. Khong phu thupc vao kim loai lam Catot cua te bao quang dien D. Ti le vdi cUdng dp ciia ehum sang kich thich
Cau 12. Khi c6 hien tupng quang dien xay ra trong te bao quang dien, phat bieu nao sau day khong diing?
A. Giii nguyen cUdng dp chiim sang kich thich va kim loai diing lam Catot, giam tan so' ciia anh sang kich thich thi dong nSng ban dau cUc dai cua eleetron quang dien tSng
B. Giu" nguyen ehCim sang kich thich, thay doi kim loai lam Cato't t h i dong nSng ban dau eUc dai eua eleetron quang dien thay doi
C. GiOr nguyen tan so' cua anh sang kich thich va kim loai lam Cato't, tang CUdng dp chum sang kich thich thi dong nSng ban dau eUc dai ciia electron quang dien khong doi
D. GiOf nguyen cUdng dp ehum sang kich thich va kim loai dung lam Catot.
giam budc s6ng ciia anh sdng kich thich thi dong nSng ban dau eye dai eua eleetron quang dien tSng
Cau 13. Trong hien tupng quang dien, van to'e ban dau eua eae eleetron quang dien bi biie ra khoi be mat kim loai se
A. CO gia t r i phu thupc vao eUdng dp eua dnh sdng chieu vao kim loai d6 B. khong phu thupc vao budc song anh sang kich thich
C. khong phu thupc vao ban cha't kim loai
D. phu thupc vao burdc song anh sang kich thich va ban chat cua kim loai lam Cato't
Cau 14. Mot nguon anh sdng phdt ra anh sang c6 tan so' f. NSng lupng mpt photon ciia anh sang nay
A. Ti le nghich v6i tan so f B. Ti le thuan vdi tan so f
C. Ti le thuan vdi cSn bac 2 eua tan so f D. Ti le nghich vdi can bac 2 ciia tan so' f .
Cau 15. Phat bieu nao sau day sai khi noi ve thuyet lufpng tilr anh sdng?
A. Thuyet lifpng tuf la cO sd de giai thich cac dinh luat quang dien.
B. Anh sang dupe tao thanh bdi edc hat gpi Ik photon
C. Nang iMng cua moi photon ciia mpt chum anh sang don s^c t i le thuan v<Ji budc song eiia ehum sang do
D. Nang lupng cua moi photon cua mpt ehum anh sang don s^e t i Ip thuan vdi tan sd eua ehCim sang do
Cau 16. Vdi e la van toe anh sang trong chan khong, f la tan so', X Ik h\idc sdng anh sang, h la hkng sd PlSng, phat bieu nao sau day sai khi ndi ve thuyet lupng tii anh sang
A. Chum anh sang la mpt chum hat, moi hat gpi la mpt photon B. Van td'c ciia photon trong chan khong la c = 3. 10^ m/s
he C. Moi mpt lu'png tijT anh sang mang nang iMng xae dinh ed gia t r i s = —
X D. Moi mpt lifpng tut dnh sdng mang nSng lupng xae dinh ed gid t r i e = h/f Cau 17. Theo thuyet lu'png tijr dnh sdng thi ndng lupng ciia
A. mpt photon ed ndng lupng t i le thuan vdi bude sdng dnh sang tiTcrng iJng vdi photon dd
B. mpt photon t i le nghich vdi tan so' cua photon dd
C. mpt photon khong phu thupc vao khoang each tCr photon dd tdi nguon phat ra no
D. eae photon trong ehum sang dcfn s^e thi khae nhau.
Cau 18. Phat bieu nao sau day sai, khi ndi ve mau nguyen tiir Bo?
A. Nguyen tijf chi ton tai d mpt so' trang thai ed ndng lu'png xae dinh, gpi la eae trang thai difng
B. Trong trang thdi dCfng, nguyen tur khong biJc xa.
C. Khi nguyen til chuyen tCr trang thdi dCfng ed ndng lupng thap E m sang trang thai dCtog ed ndng lupng eao £„ thi nguyen tuf hap thu mpt photon
CO ndng lUdng e = hfnm =
D. Trong trang thdi dCfng, nguyen tijf cd bUc xa.
Su 19. Phat bieu nao sau day sai khi ndi ve photon?
A. Van td'c ciia cde photon trong chan khong la 3. lO^m/s B. Moi photon mang mpt ndng luong xdc dinh
C. Cac p h o t o n c u a c u n g m o t a n h s d n g d o n s^c t h i m a n g c u n g m o t g i d t r j n d n g l i r o n g
D . T a n so cua m o i p h o t o n cua cdc d n h s d n g dctn s^c k h a c n h a u l u 6 n b k n g n h a u C a u 2 0 . P h a t b i e u n a o sau d a y s a i k h i n o i v l m a u n g u y e n tuf Bo?
A . T r a n g t h a i k i c h t h i c h c6 n f i n g l U O n g c a n g cao t h i b a n k i n h q u i dao cijg e l e c t r o n c a n g I d n
B . K h i or t r a n g t h a i c a b a n , n g u y e n ttir c6 n a n g l u o n g t h a p n h a t
C. N g u y e n tijT bufc xa k h i c h u y e n tii t r a n g t h d i k i c h t h i c h ve t r a n g t h a i ccf b a n D . T r o n g cac t r a n g t h a i ditog, dong n a n g cua electron t r o n g nguyen td b a n g k h o n g C a u 2 1 . V d i E I £2, E3 I a n l i r g t l a n a n g li^dng c i i a p h o t o n Ung v d i cac biJc x a m n
luc, t i a g a m m a biJc x a h o n g n g o a i t h i
A . e, > Ga > £3 B . £ 2 > E , > E3 C. £ 2 > £3 > E l D . £3 > E l > £2
C a u 2 2 . T r o n g q u a n g p h o v a c h p h a t x a ciia n g u y e n tiJf h y d r o ( H ) , d a y L a i m a n c6 A . B o n v a c h t h u p c v u n g a n h s a n g n h i n t h a y l a H „ , H p , Hg cac v a c h con
l a i t h u o c v u n g h o n g n g o a i
B . T a t ca cac v a c h deu n i m t r o n g v u n g h o n g n g o a i C. T a t ca cac v a c h d e u n ^ m t r o n g v u n g tiir n g o a i
D . B o n v a c h t h u p c v u n g a n h s a n g n h i n t h a y l a H „ , Hp, H ^ , Hg cac v a c h con l a i t h u p c v u n g ttir n g o a i
C a u 2 3 . K h i n o i ve t h u y e t l u p n g t t f a n h sang, p h a t b i e u nao dudi d a y k h o n g diing?
A Khi a n h s d n g t r u y e n d i , l i f p n g tiir a n h s a n g hi t h a y d d i v a p h u thuoc k h o a n g each t d i n g u o n s a n g
B . N g u y e n tijf h a y p h a n tuf v a t c h a t k h o n g h a p t h u h a y biJc x a a n h s a n g m o t each l i e n t u c m a t h a n h t i r n g p h a n r i e n g b i e t , d i i t q u a n g
C. N a n g l u g n g cua l u p n g t u a n h s a n g do n h o h o n n a n g lifcfng cua lucfng ti(
a n h s a n g t i m
D . M o i c h u m s a n g d u ra't y e u c u n g chufa m o t so r a t I d n lucfng tijr d n h s a n g C a u 2 4 . K h i n o i ve t h u y e t p h o t o n a n h s a n g ( t h u y e t lufcfng tijf a n h s a n g ) , p h i i t
b i e u n a o sau d a y k h o n g d u n g ?
A . V d i m o i a n h s a n g d o n s^c c6 t a n so' f xac d i n h t h i cac p h o t o n i J n g v d i aiiH s a n g do d e u c6 t a n so n h u ' n h a u
B . Budc s o n g cua a n h s a n g c a n g I d n t h i n a n g l u p n g p h o t o n t i n g v d i a n h sang do c a n g n h o
C. T r o n g c h a n k h o n g , v a n t o e c u a p h o t o n l u o n hhng v a n toe ciia a n h s a n g . D . T a n so a n h s a n g c a n g I d n t h i n S n g iMng cua p h o t o n iJng v d i a n h s a n g do
c a n g n h o
C a u 2 5 . P h a t b i e u n a o sau d a y s a i k h i n o i ve t h u y e t p h o t o n a n h s a n g ( t h u y e t l u a n g tilf a n h s a n g )
A . T a n so a n h s a n g c a n g I d n t h i n S n g l u p n g cua p h o t o n li-ng v d i a n h s a n g do c a n g lorn
B . T r o n g c h a n k h o n g , a n h s a n g c6 biTdc s o n g c a n g n h o t h i i i a n g l i f o n g cui^
p h o t o n ufng v d i a n h s a n g do c a n g I d n
C . N a n g lUOng c i i a p h o t o n t r o n g c h u m s a n g p h u t h u p c v ^ o t i n so a n h s a n g d6 D . T a n s o a n h s a n g c a n g I d n t h i n S n g l u p n g ciia p h o t o n iJfng v d i a n h s a n g do
c a n g n h o
C a u 2 6 . T h e o cac t i e n de cua B o ve cau t a o n g u y e n t\i, k h i c h u y e n tit t r a n g t h a i dCrng CO n a n g l u p n g E„ s a n g t r a n g t h a i dCfng c6 n a n g l u g n g cao h o n t h i h a p t h u m o t p h o t o n ed n a n g l u p n g b k n g
A . E „ / E „ B . E „ . E „ C . ( E n - E „ , ) D . (E^ + E J C a u 2 7 . T h e o t i e n d l t r a n g t h a i dCfng cua B o , p h d t b i e u n a o sau d a y d u n g ?
A . B i n h t h u d n g , n g u y e n t i i d t r a n g t h a i dCmg c d n a n g l u p n g t h a p n h a t g o i l a t r a n g t h a i k i c h t h i c h
B . 0 t r a n g t h a i di^ng, n g u y e n t t f l u o n buTc x a do e l e c t r o n l u o n . c h u y e n d o n g q u a n h h a t n h a n
C. N a n g l a p n g d cac m d c k i c h t h i c h l u o n I d n h o n n a n g l U P n g d mufc c P b a n D . N g u y e n tuf k h o n g t o n t a i t r o n g n h u f n g t r a n g t h a i cd n a n g li^Ong xac d i n h
g p i l a t r a n g t h a i dCrng
C a u 2 8 . T r o n g q u a n g p h o , v a c h phat x a cija n g u y e n tiJf h y d r o ( H ) , d a y B a n m e cd A . T a t ca c d c v a c h d e u n a m t r o n g vCmg s o n g v 6 t u y e n
B. B o n v a c h t h u p c v u n g a n h s a n g n h i n t h a y H „ , Hp, Hv, Hg, cac v a c h c o n l a i t h u p c v i i n g t i a R o n g h e n
C . T a t ca cac v a c h d e u nkm t r o n g v u n g tuf n g o a i .
D . B o n v a c h t h u p c v i i n g a n h s a n g n h i n t h a y l a H „ , Hp, H ^ Hg, cac v a c h c 6 n l a i t h u p c v i i n g tuf n g o a i .
C a u 2 9 . T r o n g q u a n g p h o cua n g u y e n tuf h i d r o , n e u b i e t biidc s o n g d a i n h a t c i i a v a c h q u a n g p h o t r o n g d a y L a i - m a n l a v a biidc s o n g ciia v a c h k e v d i n d t r o n g d a y n a y l a X2 t h i biidc s o n g K cua v a c h q u a n g p h o H „ t r o n g d a y
! B a n m e l a
A . A , - I2 B . C. + X2 D . —
+X2 A.2
; C a u 3 0 . T r o n g n g u y e n t i i h i d r o , b a n k i n h B o l a r,, = 5, 3. I Q - ^ ' m . B a n k i n h quy dao k i c h t h i c h thu" n h a t l a
A . 4 7 , 7 . 1 0 - " m B . 8 4 , 8 . 1 0 - ' ' m C. 2 1 , 2 . 1 0 - i ' m D . 1 3 2 , 5 . 1 0 - " m C a u 3 1 . G p i : ( I ) : Co t i n h d o n s^c cao
( I I ) : Cd t i n h d i n h h u d n g cao
( I I I ) : Cd m a t dp c o n g s u a t I d n ( c i i d n g dp m a n h ) ( I V ) : K h o n g b i k h i i c x a k h i di qua i S n g k i n h Dae d i e m n a o sau d a y k h o n g p h a i ciia t i a laze?
A . ' ( I ) B . ( I I ) can) D . ( I V )
"^au 3 2 . C h i e u t i a tuf n g o a i v a o m o t c h a t l o n g t h i c h a t n a y p h a t r a a n h s a n g m a u luc. H i e n t i f o n g n a y l a h i e n t u p n g
A . Q u a n g d i e n B . H o q u a n g d i e n C. P h a t q u a n g D . G i a o t h o a
C a u 3 3 . C h o n cau p h d t b i e u d u n g k h i n 6 i ve h u y n h q u a n g I a n q u a n g A . H u y n h q u a n g l a sir p h a t q u a n g c6 t h 6 i g i a n d a i til 10"*s t r d l e n B. L a n q u a n g Ik s y p h d t q u a n g c<5 t h d i g i a n p h a t q u a n g n g d n d u d i 10"'*s C. H u y n h q u a n g c h i x a y r a v d i c h a t l o n g
D . L a n q u a n g t h u d n g x a y r a d o i v d i c h a t r ^ n .
P H A N I I . B A I T A P T R A C N G H I E M + B A I T A P L U Y E N T A P
Van as 1: H I E N T\J0NG Q U A N G D I E N P H l / d N G P H A P
1 . N a n g laong m 5 i Ph(5t6n: (lac?ng tCf nSng lU^ng) s = h f = he
X
2. C 6 n g thQc A n h x t a n h (Einstein)
A 1 2
e = A + — m v Oniax A = he 3. Hi0u d i § n t h ^ h a m
T T 1 2 u. K= - u „ < o
e = 1,6.10-''C h = 6,625.10="' (J.s) 0 = 3.10'' (m/s) m = 9,1.10"'' k g Trong do:
* £ : N a n g iMng m o i P h o t o n (J)
* h : H ^ n g so P l a n c k (J.s)
* f: T a n so ( H z )
* X: Birdc s o n g cua d n h s d n g c h i e u v a o C a t o t ( m )
* A : C o n g t h o a t cua e l e c t r o n r a k h o i C a t o t ( J )
* XQ-. G i d i h a n q u a n g d i e n ( t h e m q u a n g d i e n ) ( m )
* m : K h o i l o t f n g e l e c t r o n ( k g )
* VQ : V a n toe b a n dau ciJc d a i cua electron k l i i t h o a t r a k h o i Catot (m/s)
* W d (, : D o n g n a n g b a n dau ciJc dai cua electron k h i thoat r a k h o i Catot (J) 4 . V$ n t d c e l e c t r o n khi d i n A n 6 t .
m v m v '
Trong do: •
*vo: V a n toe b a n d a u ciia e l e c t r o n k h i biJc r a k h o i C a t o t (m/s)
* v: V a n toe cua e l e c t r o n k h i d e n Ano't (m/s)
B A I T A P M A U
Lo^i 1: G i d l H A N Q U A N G D I E N C O N G T H O A T - N A N G L U O N G P H O T O N
B a i 1 . M o t te bao q u a n g d i f n m a k i m l o a i l a m Cato't eo g i d i h a n q u a n g d i e n l a 0,35 |im. H o i k h i c h i e u d o n g t h d i h a i bixdc s o n g k i n h t h i c h X| = 0,2 (.im v a X2 - 0,4 |im t h i c6 x a y r a h i e n tifOng q u a n g d i e n k h o n g ? V i sao?
f
Tom tat
• Xo - 0,35 n m
• Xi = 0,2 |im
• Xi = 0,4 \xm
H o i CO h i e n tiTong quang d i e n k h o n g ?
Hit&ng dan gidi
T h e o d i n h l u a t thuf n h a t t h i d i e u k i e n de x a y r a h i e n t i r g n g q u a n g d i e n \aX<XQ n e n :
• Xi <X Q x a y r a h i f n tifcfng q u a n g d i ^ n
• A.2 > ^ 0 - > k h o n g x a y r a h i e n t u o n g q u a n g d i e n
B a i 2. Ngudri t a c h i e u a n h s a n g k i c h t h i c h m a n a n g l u o n g p h o t o n l a 2 e V v a 4 eV. C a t o t l a m b a n g k i m l o a i m a c o n g t h o a t l a 2,13 eV. H o i eo x a y r a h i e n t i f o n g q u a n g d i e n k h o n g ? V i sao?
Tom tat e, = 2 eV
£ 2 = 4 eV A = 2,13 e V
H o i CO xay r a h i e n t i / o n g
• quang dien khong?
Hiidng dan gidi
T h e o d i n h l u a t q u a n g d i e n thiJ n h a t t h i de x a y r a h i f n tifcfng q u a n g d i e n p h a i t h o a d i e u k i $ n
8 > A h a y X<XQ
n e n : EJ < A - > k h o n g x a y r a h i ^ n t i / g n g q u a n g d i e n E 2 > A - > x a y r a h i e n tiTcfng q u a n g d i e n .
B a i 3 . Cato't ciia t e b a o q u a n g d i e n diTcfe l a m b ^ n g k e m eo g i d i h a n q u a n g d i e n l a 0,35 ( i m . T i n h c o n g t h o a t cua e l e c t r o n r a k h o i k i m l o a i . C h o h = i 6,625.10 e = 3.10" m/s.
Tom tat
* X Q - 0,35 ^ m T i m A = ?
h = 6,625.10-^" J.s c = 3.10® m/s
Hudng ddn gidi C o n g t h o a t cua e l e c t r o n k h o i k i m l o a i
he 6,625.10-=*^310' A = 0,35.10""
= 5 , 6 7 8 . 1 0 - ' " (J) A ô 3 , 5 4 e V
B a i 4 . H a y t i n h t a n so v a n a n g lu'cfng cua p h o t o n d n g v d i a n h s a n g t i m c6 X = 0,38 Mm. B i e t h = 6,625.10-^"* J.s, e = 3.10® m/s.
Tom tat 'X = 0,38 \xm h = 6,625.10"'''' J.s c = 3.10® m/s
Hiidng ddn gidi
• T a n so cQa, p h o t o n liTng v d i d n h s a n g t i m c_ _ 3.10"
X ~ 0 , 3 8 . 1 0 '
f = f = 7 , 8 9 4 . 1 0 " ( H z )
N a n g laang ciia photon
e = hf = he _ 6,625.10 •".3.10"
X " 0,38.10" Ê = 5,23.10 ' " J ô 3,26 eV
L o ? / 2: VAN T O C - DONG NANG C U A C A C E L E C T R O N QUANG DIE|^
B a i 1. T r o n g t h i n g h i e m ve h i e n tiTpng quang d i e n . K i m loai l a m Catot c6 - 0,55 |.im. Chieu dong t h d i h a i hxido. s o n g k i c h t h i c h c6 Xi = 0,3 |.un va X.> = 0,6 Mm.
a) H o i CO h i e n tugng quang d i e n k h o n g ? V i sao?
b) Neu CO h i e n quang d i e n t h i h a y t i n h v a n to'c ban dau ciTc d a i va dong n a n g ban dau cua cac electron k h i buTc ra k h o i Catot.
Cho h = 6,625.10'"' ( J . s ) ; c = 3.10" ( m / s ) m,. = 9,1.10 kg; e ^ 1,6.10-" (C).
Tom tat
• Xo = 0,55 |.im
• Xi = 0,3 |.im
• ^2 = 0.6 |am
a) H o i CO hien tu'png quang dien khong?
HUdng dan gidi
a) Theo d i n h luat quang d i e n thu" n h a t t h i de cc h i e n tUcfng quang d i e n c a n t h o a dieu k i e n
X < X ( ,
n e n • Xi < X n x a y r a h i e n tu'ong quang dien.
• >.2 > khong xay r a hien tiTcmg quang dien b) V,)„,„., = ? W,,max = ?
b) Van toe ban dau cue dai cua electron k h i biJc ra k h o i Catot.
Nhqn xct: Trong 2 biTdc song kich t h i c h Xi va X2 t h i chi c6 Xi xay ra hier tu'png quang dien nen a cau b nay ta chi sijf dung ^1
• Ap dung cong thuTc A n h x t a n h : E = A + he
mvn
he mv
— + O n i a x 2
V m V„ 0,813.10" n i / Dong nang ban dau ciTe dai cua cac electron k h i biirc ra k h o i Catot
W, = = 3,0074.10"'' (J) W, = 1,879625 (eV)
B a i 2. Trong t h i nghiem ve hi§n tupng quang dien. Anh sang kich t h i c h co nang lifpng photon la 3 , 2 6 eV. K i m loai l a m Catot co gidi han quang dieU la 0 , 5 7 8 nm. Hieu dien the giuTa Ano't va Catot ciia te bao quang dien 1'^
UAK = 2 5 V. T i m
a) Van toe ban dau eUc dai cua electron k h i biJc ra k h o i Cato't.
b) Van toe va dong nang cua electron k h i dap vao Anot.
Cho h = 6 , 6 2 5 . 1 0 " ' ' J.t,; c = 3.10** m/s
Electron co m„ = 9.1.lO"'^' k g ; e ^ 1 , 6 . 1 0 ' ' " C. ^ Tom tat . £ = 3 , 2 6 eV
= 5 , 2 1 6 . 1 0 " ' " (J)
• X,, = 0 , 5 7 8 \xm := 0.578.10"*^ m . U A K = 2 5 V
. h = 6 , 6 2 5 . 1 0 " ^ ' J.s
• e = 3.10** m/s
• me = 9 , 1 . 1 0 " ' " k g . e = 1 , 6 . 1 0 " ' ' C a) T i m Vo^a. = ?
b) V = ?
W„ = ?
HU&ng dan gidi a) T i m Vo^a^ = ?
Ap dung cong thiifc A n h x t a n h
£ = A . • - he 2
= 0,625.10''(m/s)
b) V a n to'c ciia electron k h i dap vao Ano't Ap dung cong thufe
B.UAK = mv m v '
2 2 Dpng nang ciia electron k h i den A n o t
V = 3,03.10" m / s
mv = 4.177.10""* (J) h a y W„ = 26.1 eV
Lo^i 3: HIEU DIEN T H E HAM
B a i 1. Trong t h i nghiem ve h i e n tiTpng quang dien. Anh sang kich t h i c h co tan so 6.10''* Hz. Catot ciia te bao quang dien co gidi han quang dien la 0,75 \i.m..
Cho h = 6,625.10-''^ J.s, e = 3.10" m/s, m , = 9,1.10"'" kg; -e = -1,6.10"" c.
a) T i n h cong thoat eiia electron k h o i k i m loai l a m Catot.
b) T i m dp Idn h i e n dien the h a m de t r i e t tieu dong quang dien.
c) T i m dieu k i e n hieu dien the UAK de t r i e t tieu dong quang dien.
Tom tat
• f = 6.10'" Hz
• Xo - 0,75 |iim
• h = 6,625.10"'"' J.s
• c = 3.10" m/s
• m„ = 9,1.lO"'^' k g
• e = 1,6.10"" (C) a) T i m A = ?
ci UAK the nao de I = 0
lliidng dan gidi
a) Cong thoat ciia electron k h o i k i m loai l ^ m Catot A = — = 2 , 6 5 . 1 0 " " J
~Xa
h a y A = 1,65625 eV
b) Dp Idn hieu dien the h a m de t r i p t tieu dong dien Ap dung cong thiJc A n h x t a n h
mv
E = A + (1) ma m v :
(2) The (2) v a o ( l )
( l ) - > h f = A + e.Ui, <->Uh = h f - A
^ U h = 6,625.10 ".6.10'^ -2,65.10 1,6.10"
<-> U|, = 0,828 (V)
c) B i l u k i e n UAK de t r i e t tieu dong quang dien UAK ^ - U h U - K < -0,828 (V)
Hinh ve rninh hga
I ( A )
UAK S - U h thi I = 0 Ibi,/
U A K ( V )
- 0 . 8 2 8 ( v ) O
B a i 2. Trong t h i nghiem ve hien tirgng quang dien. Ngiidi ta thay rkng k h i U^K
< - 2 V t h i dong quang dien triet tieu. T i m van toe cua electron den Anot k h i hieu dien the UAK ^ +2 V. Cho m,, = 9,1.10"'^ kg, e = 1,6.10-^" (C).
Tom tat
. U A K ^ - 2 V . U A K = +2 V
• me = 9,1.10"^' k g . e = l.e.lO"''' C.
T i m V = ?
Hudng dan gidi
• Dieu k i e n UAK de t r i e t tieu dong tieu UAK ^ - U h .
• De cho U A K < - 2 V
^ U„ = 2 V.
• V a n to'c electron k h i den Anot
,2
e . U A K = mv mvn mv
M a e.Uh = 22122^
(1) (2)
The (2) vao (1) ^ B.UAK = mv - e . U h
^ v = / e ( U , , + U „ ) . - ^ m
v = 1,185.10" m/s
BAI TAP TRAC NGHIEM
C a u 1. Hay t i n h t a n so', nSng luong cua photon iJng v d i a n h sang do co I = 0,76 (nm).
A. 3,9.10^* Hz; 1,63 eV B . 5.10^" Hz; 2 eV C. 6.10'''Hz; 6 eV D. 7.10'" Hz; 7 eV.
C a u 2. M o t te bao quang dien co Catot hKng k i m loai c6 gidi h a n quang dien X,) = 0,578 (^im). Cong thoat cua k i m loai la
A. 1 eV B . 2,14 eV C. 2 eV D. 3 eV
C a u 3. Cho cong thoat cua k i m loai dung lam Catot cua te bao quang dien la 3 eV Chieu tdi Catot li = 0,4 (|.im), X2 = 0,45 (^m).
A. Xi xay ra, X2 k h o n g xay ra B . X2 xay ra, khong xay ra C. Xi, A.2 xay ra D. Xi, X2 khong xay ra.
C a u 4. Cho cong thoat cija k i m loai dung l a m Catot cua te hko quang d i f n 1^
2,4 eV. Chieu t d i Catot Xi = 0,66 (^m), X2 = 0,45 ((.tm).
A. X]_ xay ra, X2 k h o n g xay ra B . Xi, X2 deu xay ra.
C. X2 khong xay ra, xay ra. D. Xi, X2 khong xay ra.
C a u 5. Gidi h a n quang dien cua N a t r i la 0,5 ((im). Chieu vao N a t r i t i a tijf ngoai CO = 0,25 (urn). Cho h = 6,625.10""'"' J.s; c = 3.10** m/s; m = 9,1.10-'"kg;
e = 1,6.10""^C.Van toe cifc dai ciia cac electron k h i bufc ra k h o i Catot la A. 934679 m/s B . 9 . 1 0 ' m / s C. 8.10*^ m/s D. 9.10^ m/s
C a u 6. Gidi han quang dien ciia k i m loai lam Catot la 0,56 (^m). Ngirdi ta chieu X = 0,45 (|.im) vao Catot. Cho h = 6,625.10"'^'' J.s; c = 3.10** m/s; m = 9,1.10"^'kg;
e = 1,6.10"'^C. Dong nang ban dau circ dai cua electron k h i biJc ra khoi Catot la A. 0 , 2 e V B . 0 , 4 e V C. 0 , 5 4 e V D. 0,8 eV
ZsiVL 7. Gidi h a n quang dien eiia k i m loai l a m Catot la 0,6 ((.im), chieu a n h sang kieh thich co X ^ 0,4 (pm). Cho h = 6,625.10-''"J.s; c = 3.10** m/s; m = 9,1.10~'"kg;
e = 1,6.10"'''C . Do Idn ciia hieu dien the h a m la
A, 1,03 V B . 3,05 V C. 2 V D. 4,03 V
C a u 8. Chieu t d i te bao quang dien a n h sang co bu'dc song 4 2 0 0 ^ " , gidi h a n quang dien la 5000 A . Cho h = 6,625.10"^" J.s; e = 3.10** m/s; m = 9,1.10-^'kg;
e = 1,6.10"'^C. T i m dieu k i e n hieu dien the giiJa Anot va Cato't de t r i e t tieu dong dien.
A. U A K < -0,2 V B . U A K > - 0 , 2 V C. U A K < -0,47 V D. U A K > -0,47 V
o
C a u 9. Chieu t d i te bao quang dien anh sang co bifdc song X = 4500 A va Cato't CO gidi han A,, = 6000 A . Cho h = 6,625. lO"'''' J.s; e = 3.10** m/s; m = 9,1.10~^'kg; e
= 1,6.10"'^C, UAK = 40 (V). V a n toe electron k h i den Anot la I A. 3,78.10'm/s B . 3,78.10^ nVs
C. 3,78.10^ m/s D. 3,78.10* m/s
C a u 10. Cho cong thoat ciia electron k h o i k i m loai la 4 (eV). Chieu a n h sang c6 X = 2500 A vao Catot. Cho h = 6,625.10"^" J.s; e = 3.10® m/s; m = 9,1.10"^'kg;
e = 1,6.10"'^C, UAK = 100 (V). Dong nang ciia electron k h i den A n o t 1^
A. 100,9 eV B.10,09 eV C. 1,009 eV D. 0,1009 J .
C a u 11. K h i chieu a n h sang X = 0,45 ((.im) vao Cato't cua te bao quang dien t h i van to'c ban dau cUc dai ciia electron thoat ra k h o i Cato't la 0,5. l O ' (m/s). Cho I h = 6,625.10"'" J.s; c = 3.10* m/s; m = 9,1.10 kg; e = 1,6.10-"C. Gidi h a n
quang dien cua k i m loai l a m Catot la
A. 0,6 (|.tm) B . 0,5 (^m) C. 0,7 (pm) D. 0,75 (nm).