MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 11
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Q.1
The photoelectric work function for a metal surface is 4.125eV. The cut off wavelength for this surface is . [ CBSE-PMT 1999]
0%
a) 4125 Å
0%
b) 3000 Å
0%
c)6000 Å
0%
d)2062.5 Å
Explanation
let λ0 be cut off wave lengthWork function Φ=hc/ λ0 ∴ λ0=hc/ Φ here Φ=4.125×1.6×10⁻¹⁹h=6.26×10⁻³⁴c=3×108 Substituting values in above equation on solving we getλ=3000Å Answer:(b)
Q.2
The momentum of photon of energy 1 MeV in kg m/s, will be [ CBSE-PMT 2006]
0%
b) 10-22
0%
d) 0.33×106
0%
a) 7×10⁻²⁴
0%
c) 5×10⁻²²
Explanation
1MeV=106×1.6×10⁻¹⁹ Momentu of photon=E/c Momentum of photon=1.6×10⁻¹³ / 3×108 Momentum of photon=5×10⁻²² kg m/sec Answer: (c)
Q.3
The wave length associated with an electron accelerated through a potential difference of 100V, is of the order of .. [ CBSE-PMT 1996]
0%
a) 1000 Å
0%
b) 100 Å
0%
c)10.5 Å
0%
d)1.2 Å
Explanation
Potential difference V=100V. Kinetic energy=P.E Kinetic energy=qV Kinetic energy=1.6×10⁻¹⁹ ×100=1.6×10⁻¹⁷ J now energy E=hc/ λ ∴ λ=hc / E Answer: (d)
Q.4
Photoelectric work function of a metal is 1eV. Light of wave length λ=3000 Å falls on it. The photo electron come out with velocity [ CBSE-PMT 1991]
0%
a) 10 m/sec
0%
b) 102 m/sec
0%
c) 104 m/sec
0%
d) 106
Explanation
We know that hν=Φ + ½ ( m v2) Φ=1eV=1× 1.6×10⁻¹⁹ J Answer: (d)
Q.5
The wave length of 1keV photon is 1.24×10⁻⁹m. What is the frequency of 1MeV photon?
0%
a)1.24×1015
0%
b) 2.4×1020
0%
c)1.24×1018
0%
d)2.4×1023
Explanation
Here hν=103 eV and λ=1.24×10⁻⁹m.Now ν=c/λ ν=3×108 / 1.24×10⁻⁹ν=2.4×1017∴ frequency of 106 will be Thus frequency of 106 is 2.4×1020 Answer: (b)
Q.6
Kinetic energy of an electron, which is accelerated in a potential difference of 100V is [ CBSE - PMT 1995]
0%
a) 1.6×10⁻¹⁷J
0%
b) 1.6×10⁻¹⁹J
0%
c)1.6×10⁻²¹
0%
d)1.6×10⁻²⁵J
Explanation
Kinetic energy=qE Kinetic energy=1.6×10⁻¹⁹×100 Answer:(a)
Q.7
An electromagnetic radiation of frequency 3×1015 c/s fall on a photoelectric surface whose work function is 4.0eV. The maximum velocity of photoelectrons emitted by the surface is ( h=6.6×10⁻³⁴ Js) [ JIMPER 19989]
0%
a)1.73×106 m/sec
0%
b) 1.73×105 m/sec
0%
c)1.73×104 m/sec
0%
d)1.73×103 m/sec
Explanation
Energy of incident radiation E=hν E=6.6×10⁻³⁴×3×1015E=19.8×10⁻¹⁹ J Work function Φ=4.0eV=4×10⁻¹⁹JEnergy of emitted electron=E-ΦEnergy of electron=19.8×10⁻¹⁹ - 4×10⁻¹⁹Energy of Electron=15.8×10⁻¹⁹Kinetic energy of electron ½ mv2=15.8×10⁻¹⁹½ ×9.11×10⁻³¹=15.8×10⁻¹⁹v2=3.468×1012v=1.73×106Answer: (a)
Q.8
Given h=6.6×10⁻¹⁴ joules×sec. The momentum of each photon in give radiation is 3.3×10⁻²⁹ kg×metre (sec)-The frequency of radiation is [ CBSE 1990]
0%
a)3×10⁻² hz
0%
b) 6×1010
0%
c)7.5×1012 hz
0%
d)1.5×1013 hz
Explanation
λ=h/p and λ=c/νfrom above two equations ν=pc/λν=(3.3×10⁻²⁹×3×108 ) / 6.6×10⁻³⁴ν=1.5×1013 hzAnswer: (d)
Q.9
The energy of photon is 10eV. The momentum of a photon is
0%
a) 5.33×10⁻²⁵ kg×m /s
0%
b) 5.33×10⁻²⁷ kg×m /s
0%
c)5.33×10⁻²⁹ kg×m /s
0%
d)5.33×10⁻²³ kg×m /s
Explanation
Energy E=mc2 momentum p=mc Thus E=pc or p=E/cp=10×1.6×10⁻¹⁹ / 3×108 p=5.33×10⁻²⁷ kg×m / secAnswer: (b)
Q.10
When the light of wavelength 2537Å is made incident over the surface of copper slab, the stopping voltage 0.24 volt obtained. The threshold frequency for copper will be .. [ Raj.PMT 1996]
0%
a) 1.124×1015 Hz
0%
b) 1.414×1014 Hz
0%
c)2.248×1015 Hz
0%
d)None of the above
Explanation
incident energy=work function + maximum kinetic energy incident energy=hc/λ Incident energy=6.6×10⁻³⁴ ×3×108 / 2537×10⁻¹⁰ Incident energy=0.0078×10⁻¹⁶maximum kinetic energy=eV=1.6×10⁻¹⁹×0.24=0.384×10⁻¹⁹Now Work function=Incident energy - maximum kinetic energyWork function=0.0078×10⁻¹⁶ - 0.384×10⁻¹⁹Work function=0.007416×10⁻¹⁶Work function=h×νo h×νo=0.007416×10⁻¹⁶νo=0.007416×10⁻¹⁶ / 6.6×10⁻³⁴νo=1.124×1015 Answer:(a)
Q.11
Given h=6.6×10⁻³⁴ joules×sec. the momentum of each photon in a given radiation is 3.3×10⁻²⁹ kg×metre / sec .. [ CBSE 1990]
0%
b) 6×1010 hz
0%
c) 7.5×1012 hz
0%
d) 1.5×1013hz
0%
a) 3×10⁻² hz
Explanation
λ=h/p and λ=c/ν c/ν=h/p ν=pc/λ ν=3.3×10⁻²⁹ ×3×108 / 6.6 ×10⁻³⁴ ν=1.5×1013 hz Answer: (d)
Q.12
When electrons are accelerated through potential difference of V volts, the de Broglie wavelength associated is given by
0%
a) √(150/V) Å
0%
b) √(150/V) m
0%
c)(150/ √V) Å
0%
d)(√150/V) Å
Explanation
λ=h / √(2meV)on substituting values of h, m and e we get λ=12.3×10⁻¹⁰ / √V metreλ=12.3/ √V ÅOr λ=√(150/V) Å Answer:(a)
Q.13
The de-Broglie wavelength of an electron moving with speed of 6.6×105 m/s is approximately [ CPMT 1993]
0%
a) 10-11 m
0%
b) 10-9 m
0%
c)10-7 m
0%
d)10-5 m
Explanation
use formula λ=h/mv h=6.6×10⁻³⁴ and m=9.1 ×10⁻³¹ kg Answer:(b)
Q.14
The wavelength of most energetic X-rays emitted when a metal target is bombarded by 40 KeV electrons, is approximately... [MPPMT 1999]
0%
a) 300 Å
0%
b)10 Å
0%
c)4 Å
0%
d)0.31 Å
Explanation
USe formula λ=ch/eV 40 KeV=40×103×1.6×10⁻¹⁹ Answer:(d)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)