Daily Practice Problem Sheet 115

TOPIC - MODERN PHYSICS

1. Let nr and nb be respectively the number of i i
photons emitted by a red bulb and a blue bulb of
equal power in a given time.
(A) nr = nb (B) nr < nb (C) (D)
(C) nr > nb (D) data insufficient
t t
2. 10 –3 W of 5000 Å light is directed on a
photoelectric cell. If the current in the cell is 8. The maximum kinetic energy of photoelectrons
0.16 mA, the percentage of incident photons emitted from a surface when photons of energy
which produce photoelectrons, is 6 eV fall on it is 4eV. The stopping potential is
(A) 0.4% (B) .04% Volts is :
(C) 20% (D) 10% (A) 2 (B) 4 (C) 6 (D) 10

3. If the frequency of light in a photoelectric 9. Radiation of two photon energies twice and five
experiment is doubled, the stopping potential will times the work function of metal are incident
(A) be doubled sucessively on the metal surface. The ratio of
(B) halved the maximum velocity of photoelectrons emitted
(C) become more than doubled is the two cases will be
(D) become less than double (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

4. The stopping potential for the photo electrons 10. Cut off potentials for a metal in photoelectric
emitted from a metal surface of work function effect for light of wavelength 1, 2 and 3 is
1.7eV is 10.4 V. Identify the energy levels found to be V1, V2 and V3 volts if V1, V2 and V3
corresponding to the transitions in hydrogen are in Arithmetic Progression and 1, 2 and 3 will
atom which will result in emission of wavelength be :
equal to that of incident radiation for the above (A) Arithmetic Progression
photoelectric effect (B) Geometric Progression
(A) n = 3 to 1 (B) n = 3 to 2 (C) Harmonic Progression
(C) n = 2 to 1 (D) n = 4 to 1 (D) None
5. When a photon of light collides with a metal 11. Photons with energy 5eV are incident on a
surface, number of electrons, (if any) coming cathode C, on a photoelectric cell. The maximum
out is energy of the emitted photoelectrons is 2eV.
(A) only one (B) only two When photons of energy 6eV are incident on C,
(C) infinite (D) depends upon factors no photoelectrons will reach the anode A if the
stopping potential of A relative to C is
6. A point source of light is used in photoelectric
(A) 3V (B) –3V (C) –1V (D) 4V
effect. If the source is removed farther from
the emitting metal, the stopping potential : 12. By increasing the intensity of incident light
(A) will increase keeping frequency (v > v0) fixed on the surface
(B) will decrease of metal
(C) will remain constant (A) kinetic energy of the photoelectrons increases
(D) will either increase or decrease (B) number of emitted electrons increases
(C) kinetic energy and number of electrons
7. A point source causes photoelectric effect from
increases
a small metal plate. Which of the following curves
(D) no effect
may represent the saturation photocurrent as a
function of the distance between the source 13. In a photoelectric experiment, electrons are
and the metal? ejected from metals X and Y by light of intensity
i I and frequency f. The potential difference V
i
required to stop the electrons is measured for
various frequencies. If Y has a greater work
function than X ; which one of the following
(A) (B)

t t
 graphs best illustrates the expected results ? 17. If h is Planck’s constant is SI system, the
v v momentum of a photon of wavelength 0.01 Å is:
(A) 10–2 h (B) h (C) 102 h (D) 1012h
X X
(A) (B) 18. The angular momentum of an electron in the
Y Y 3h
o f o f hydrogen atom is . Here h is Planck’s
2
v v constant. The kinetic energy of this electron is :
(A) 4.53 eV (B) 1.51 eV
X Y (C) 3.4 eV (D) 6.8 eV
(C) Y (D) X
19. Consider the following electronic energy level
o f o f diagram of H-atom : Photons associated with
shortest and longest wavelengths would be
14. A image of the sun of formed by a lens of focal- emitted from the atom by the transitions labelled.
length of 30 cm on the metal surface of a photo-
electric cell and a photo-electrci current is
produced. The lens forming the image is then n=
A
replaced by another of the same diameter but of n=4
focal length 15 cm. The photo-electric current D
C n=3
in this case is
(A) I/2 (B) I (C) 2I (D) 4I B
n=2
15. A proton and an electron are accelerated by
same potential difference have de-Broglie n=1
wavelength p and e. (A) D and C respectively
(A) e = p (B) e < p (B) C and A respectively
(C) e > p (D) none of these (C) C and D respectively
16. An electron with initial kinetic energy of 100eV (D) A and C respectively
is acceleration through a potential difference of 20. If the electron in a hydrogen atom were in the
50V. Now the de-Broglie wavelength of electron energy level with n = 3, how much energy in
becomes. joule would be required to ionise the atom ?
(A) 1 Å (B) 1. 5 Å (Ionisation energy of H-atom is 2.18 × 10–18 J):
(A) 6.54 × 10–19 (B) 1.43 × 10–19
(C) 3Å (D) 12.27 Å –19
(C) 2.42 × 10 (D) 3.14 × 10–20

Name ________________________ Date - 18-12-22

DPP- 115 WAVE OPTICS Mobile No. ________________

AB C D AB C D AB C D AB C D
1. O O O O 6. O O O O 11. O O O O 16. O O O O
2. O O O O 7. O O O O 12. O O O O 17. O O O O
3. O O O O 8. O O O O 13. O O O O 18. O O O O
4. O O O O 9. O O O O 14. O O O O 19. O O O O
5. O O O O 10. O O O O 15. O O O O 20. O O O O
 Daily Practice Problem Sheet 115
TOPIC - MODERN PHYSICS
1. C Max KE = 4ev
 = Ephoton - Kmax
12400(eV)
E= = 6 - 4 = 2v
λ(in Å) then stopping Potential is 4v.
IAt λ 9. A
No. of Photon=
hc
1 2
2   K1  K1   mv1
Pt λ Eλ 2
No. of Photon= =
hc hc
1 2
if E is constant no. of photon is   5   K 2  K 2  4  mv2
2
2. B
v1 :v2 1 : 2
-3
10
No.of Photons= 10.C
12400 -13
×1.6×10 If v1, v2, v3 are in A.P. then
5000
16 hc
=0.25×10 = +eV
λ 1 ....(1)
-6 1
- 0.16×10 +12
No.of e reaching= =10
-19 hc
1.6×10 = +eV
λ 2 ....(2)
12 2
10
%= ×100=0.04%
16
0.25×10 hc
=+eV
λ 3 ....(3)
3. C 3
hf = +ev s After solving (1), (2) and (3) we get

4. A 21 3
2 
1  3 which are in H.P..
hf =1.7+10.4=12.1eV=energy
in H-atom 11.B
KEmax = 2ev
x=3 Ephoton = 5ev
0 = 5ev - 2ev = 3ev
12.09 ev Now no current when
Ephoton = 6ev
n=1
i.e. KEmax < 3ev
5. A eVmax < 3ev
A Photon can interact with only a single Vmax < 3ev/e = 3V
electron.
  =3eV
6. C
 K.E.max =3eV in Second case
Depends on f not on Intensity
12.B
7. D
no. of Photons  I
As distance  ses. I , no. of photon e- ejection 
I  ses. 13.A
Greater work function means greater cut off
 i 
frequency.
P Slope Remains same
I y>x
4 r2
Intercept of y > Intercept of x
8. B and must be parallel to each
Ephoton = 6 ev
14.B
Diameter is same so light falling will be same so 18.B
photoelectric current will be same. nh
J = mvr =
15.C 2
They have same K.E.  n=3

h 1
λ= K.E. = - T.E. = 13.6 ×
2m K.E. 9
mp > me and qp=qe = 1.51ev
19.C
1
p<e as  
m  1 1  hc
 E  Rcz 2  2  2  
16.A  n1 n2  
KE = 100+50 = 150eV C 
 Shortest
v = 150volt D 
 longest
150 20.C
λ=
V 2.18  1018
I.E. 
 1 A
0
n2

17.D 2.18  1018


h
9
12
= 10 h = 2.42 × 10 -19 J
λ

ANSWER KEY
1. (C) 2. (B) 3. (C) 4. (A) 5. (A) 6. (C)

7. (D) 8. (B) 9. (A) 10. (C) 11. (B) 12. (B)

13. (A) 14. (B) 15. (C) 16. (A) 17. (D) 18. (B)

19. (C) 20. (C)