1. E = hn۪
E ∝ 1/λ energy of proton reduces to half.
2. Alkali metals have too low work functions. Even visible light can eject electrons from them.
3. UV are most effective since they have highest frequency hence more energetic.
4. Yes. X-rays cause photoelectric effect in sodium, zinc & copper.
5. K.E of photons remains unaffected since they do not depend
6. stopping potential V0 = Kmax/e = 5ev/e =5 V
7. w0 = hν0 = hc/λ0
∴λ0 ∝ 1 / w0
Since sodium has lower work functions than copper it is easier for electron ejection. As it is lower work function, higher wavelength.
8. Photocells are used for reproduction of sound.
9. 1/2 mv2 = (m2 v2 )/2m = p2 /2m
According to De Broglie wave length λ = h/p
λe/ λp = pp/pe = √(mp /me )
me < mp
electrons have greater De broglie wavelength than proton .
10. θ = 90 - Φ/2
= 90 – 52/2 =64°
11. Ε = hc / λ
= 3.3×10-19 J
12. KE of photoelectrons is given by Einstein’s photoelectric equation.
Ek =1/2 mv2
= h ט- w0
V α 1 / √λ
As wavelength decreases velocity increases.
14. E= h c/6.6) = ג x 10-34 x 3 x 108)/ (4 x 10-7) = 4.98 x 10-19 j
E= (4.98 x 10-19)/ (1.6 x 10-19) =3ev
Hence, metal x will emit electrons.
15. For a photon E1=hc/ ג
For an electron ג=h/mv or m=h/ גv
Therefore,E2>E1.thus, electron has total energy greater than that of photon.
. 16 ph= גe= ג=h/mv
K.E. of electrons E=1/2mv2=1/2 m [h/m2[ ג
17 . E=1/2 mv2=m2v2/2m=p2/2m ∴p=√2Em
From Kinetic theory of gases average K.E.=3/2 KT
ג =h/(√2m(3/2 KT)=h/(√3m KT)
18. The energy of light obtained from the bulb is much less than work function of the wodden block.Hence no photon electrons are emitted.
19. Mo will not emit photo electron , because its work function is more than 4 ev.
20. Alpha particles due to its largest mass.
21. R α 1/q
22. Ee = Mv2/r
23. λ = λ /√2.
24. λ / 2.
25. λ = hc/Φ = 2823 Aͦ .
CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)