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Physics NEET MCQ
Particle Nature Of Light Mcq
Quiz 7
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Q.1
Light of wavelength λ strikes a photo sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wave length must be changed to λ' where [ MPPMT 1997]
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a) λ'=λ/2
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b) λ'=2λ
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c)λ' > λ
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d)λ/2 < λ' < λ
Explanation
Above relations will be satisfied if λ' > λ/2 and λ' < λ Answer:(d)
Q.2
In photo emissive cell, with exiting wavelength is changed to λ/4, the speed of fastest electron will be [ CBSE PMT 1998]
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a)
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b)
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c)
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d)
Explanation
Answer: (d)
Q.3
The photoelectric work function for a metal surface is 4.125 eV. The cut off wavelength for this surface is [ CBSEPMT 1999]
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a)4125 Å
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b) 2062.5 Å
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c)3000 Å
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d)6000 Å
Explanation
Use formula φ=hc/eλ . Answer: (c)
Q.4
The slope of frequency of incident light and stopping potential for a given surface will be [MPCET 1999]
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a) h
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b) h/e
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c)eh
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d)e
Explanation
Answer: (b)
Q.5
The kinetic energy of electron moving with velocity of 4 ×106 m/s will be [ MNR 1999]
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a) 30 eV
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b)45 eV
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c)50 eV
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d)60 eV
Explanation
Answer:(d)
Q.6
The work function of any metal is 4eV. For emitting photoelectrons of zero velocity from the surface of this metal, the wavelength of incident light required must be [ MNR 1999]
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a) 2700 Å
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b) 1700 Å
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c) 5900 Å
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d) 3100 Å
Explanation
USe formula λ=hc/eφ (as work function is given in electron volts) Answer: (d)
Q.7
Who indirectly determined the mass of the electron by measuring the charge of electron [ CBSEPMT 2000]
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a)Rutherford
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b) Einstein
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c)Thomson
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d)Millikan
Explanation
Answer: (d)
Q.8
The curve drawn between velocity and frequency of photon in vacuum will be a [MPPMT 2000]
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a) straight line parallel to frequency axis
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b) straight line parallel to velocity axis
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c)straight line passing through origin and making an angle 45° with frequency axis
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d)hyperbola
Explanation
velocity of photon is independent of frequencyAnswer: (a)
Q.9
The de-Broglie wavelength of an electron in the first Bohor orbit is [ KCET 2002]
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a) equal to circumference of the first orbit
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b) equal to twice the circumference of the first orbit
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c)equal to half the circumference of the first orbit
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d)equal to one-fourth the circumference of two first orbit
Explanation
From Bohor postulate mvr=nh/2π∴ 2πr=nh/mv but h/mv=λ ∴ 2πr=n × λcircumference=n × λfor n=1circumference=λ Answer:(a)
Q.10
The de-Broglie wavelength of a particle moving with velocity 2.25 ×108 m/s is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the enrgy of the photon is [ EAMCET 2003]
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a) 1/8
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b) 3/8
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c) 5/8
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d) 7/8
Explanation
LEt m1 be the mass of particle and m2 be the equivalent mass of photon both have same de_broglie wavelength then m1v=m2c or m1 / m2=c/v=3×108/2.25×108=3/2.25 Answer: (b)
Q.11
The maximum kinetic energy of photoelectrons emitted from the surface when photons of energy 6eV fall on it is 4eV. The stopping potential in volts is [ IIT 1997]
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a)4
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b) 2
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c)6
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d)10
Explanation
Answer: (a)
Q.12
A charged oil drop falls with terminal velocity v in absence of electric field. An electric field E, keeps it stationary. The drop acquires charge q it starts moving upward with velocity v. The initial charge on the drop was
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a) q/2
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b) q
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c)2q
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d)4q
Explanation
Let initial charge be q'. given that in absence of electric field drop moves with terminal velocity therefore viscus force is acting on the drop when electric field is applied drop becomes stationary Eq'=6πηrv When drop moves upwards then Eq=6πηr(v+v) from above equations q'/q=1/2 q'=q/2Answer: (a)
Q.13
In Thomson's experiment, the same H.T. supply provides potential to anode, as also to positive deflecting plates in the region of crossed fields. If the supply voltage is doubled, then value of the new magnetic field to keep the electron beam un-deflected will be
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a) B/2
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b) √2 B
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c)B
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d)2B
Explanation
Electric fore=Eq Magnetic force=qvBTo keep electron beam un-deflected Electric force=magnetic force eE=evB B=E/v ( here v is the velocity of electron when acceleration potential is V) Velocity of electron depend on the potential Thus ½ m v2=eV v=√ (2eV/m) Thus velocity v ∝ √V ∴ B ∝ E/√V Also E=V/d E ∝ V ∴ B ∝ V/√V B ∝ √V --(1) Now potential is doubled B'∝ √2V --(2)From (1) and (2) we get B'/B=√2 B'=√2 B Answer:(b)
Q.14
A stream of electrons enters an electric field normal to the lines of force with velocity of 3 ×107 m/s. The electric intensity is 1800 V/m. The electron beam is deflected by 2 mm, while travelling through a distance of 10cm. Then e/m in will be
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a) 2 × 1014
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b) 2 × 1011
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c) 2 × 10
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d) 2 × 104
Explanation
Let electric beam enter electric field along x-axis with velocity v Therefore there no y component of velocity and acceleration along x axis displacement along x axis l=v ×t thus time taken to travel l distance is t=l/v Now initial velocity along y axis is zero and acceleration is due to electric field F=eE and F=ma ma=eE or a=eE/m displacement along y axis=(1/2) at2 since initial velocity is zero along y-axis y=(1/2) (eE/m) ( l/v)2 Answer: (b)
Q.15
A radiostation is transmitting waves of wavelength 300 m. If the radiation power of a transmitter is 10kW, then the number of photons emitted per second will be
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a)1.5 × 1029
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b) 1.5 × 1031
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c)1.5 × 1033
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d)1.5 × 1035
Explanation
Enrgy in one second=10kJ=104 and energy of each photon=hc/λenergy of n photon E=nhc/λ energy of radio transmittern=E× λ / hcAnswer: (b)
Q.16
The difference of kinetic energy of photoelectrons emitted from a surface by light of wavelength 2500 Å and 5000 Å will be
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a) 1.98 ×10-19 J
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b) 1.98 ×10-19 erg
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c)3.96 ×10-19eV
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d)3.96 ×10-19J
Explanation
use equation E=hc/λ - φ for both the wave lengths Answer: (d)
Q.17
On using light of wavelength 6000 Å, the stopping potential for a photocell is 2.4 V. If light of wavelength 4000Å is used, then stopping potential will be
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a) 1.91 V
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b) 2.91 V
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c)3.43 V
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d)4.42 V
Explanation
Use formula for stopping potential Answer:(c)
Q.18
When a piece of metal is illuminated by monochromatic light of wavelength λ then the stopping potential for photoelectric current is 3Vo. When the same surface is illuminated by light of wavelength 1.5λ, then stopping potential becomes Vo. The value of threshold wavelength for photoelectric emission will be
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a) (4/3) λ
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b) 2 λ
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c) 3 λ
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d) 4λ
Explanation
Answer: (b)
Q.19
The work-function of a substance is 4.0eV. The longest wavelength of light that can cause photoelectric emission from the substance is approximately [ IIT 1998]
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a)540 nm
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b) 400 nm
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c)310 nm
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d)220 nm
Explanation
use formula eV=hc/λ ( as energy is given in eV) Threshold wavelength is longest wavelength which can give photo emission. Answer: (c)
Q.20
The wavelength associated with a gold ball weighing 200g and moving at speed of 5m/h is of the order of ( h=6.625 × 10-34 Js) [ IIT 2001]
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a) 10-10 m
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b) 10-20 m
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c) 10-30 m
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d) 10-40 m
Explanation
use formula λ=h/mvAnswer: (c)
Q.21
The potential difference applied to an X-ray tube is 5kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is [ IIT 2002]
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a) 2 × 1016
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b) 5 × 1016
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c)1 × 1017
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d)4 × 1015
Explanation
use I=ne/t here t=1 sec Answer:(a)
Q.22
Light of frequency 1.5 times the threshold frequency is incident on photo-sensitive material. If the frequency is halved and intensity is doubled, the photo-current becomes
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a) quadrupled
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b) doubled
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c) halved
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d) zero
Explanation
threshold frequency is ν, given that incident frequency in second caase is 0.75ν thus photoelectrons will not emit Answer: (d)
Q.23
When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m for the electron is 1.8×1011 C/kg the maximum velocity of the ejected electron is [ kerala CET 2002]
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a) 6 × 105 m/s
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b) 8 × 105 m/s
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c) 1.8 × 106 m/s
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d)106 m/s
Explanation
kinetic enegy=(1/2)mv2 and maximaum kinetic energy=eV eV=(1/2) mv2 Answer: (c)
Q.24
The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal ( for which work function is 1.65 eV) will be [ JIPMER 2002]
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a)4 × 1010 Hz
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b) 4 × 1011Hz
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c)4 × 1014 Hz
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d)4 × 10-10 Hz
Explanation
use formula φ=hν ( ν is frequency) convert eV in joules and substitude for φAnswer: (c)
Q.25
When radiation of wavelength λ is incident on a metallic surface, the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double wavelength, then the stopping potential becomes 1.6V. Then threshold wavelength for the surface is [ EAMCET 2003]
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a) 3λ
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b) 4λ
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c)6λ
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d)8λ
Explanation
Answer: (b)
Q.26
The energy of photon is equal to the K.E. of a proton. The energy of the photon is E. Let λ1 be the de-Broglie wavelength of the proton and λ2 be the wavelneth of photon. the ratio of λ;1 / λ2 is proprortioanl to [ IIT 2004]
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a) Eo
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b) E1/2
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c)E-1
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d)E-2
Explanation
We know thatde-Brogli wavelength for particles in terms of energy λ1=h/ √(2mE) energy of photon E=hc/λ2λ2=hc/EBy taking ratio Answer:(b)
Q.27
If K1 and K2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength λ1 and λ2 respectively incident on a metallic surface. If λ1=3λ2 the [ EAMCET 2004]
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a) K1 ≥ K2/3
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b) K1 < K2/3
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c) 3K1=K2
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d) K1=3K2/3
Explanation
USe equation energy K=hc/λ - φ take two cases and solve to find relation between K1 and K2 Answer: (b)
Q.28
If E and p are the respective energy and momentum of photon, then on reducing the wavelength of the photon
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a)both p and E will decrease
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b) both p and E will increase
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c)p will increase but E will decrease
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d)p will decrease but E will increase
Explanation
Answer: (b)
Q.29
the momentum of photon of energy 1MeV will approximately be
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a) 10-22 kg-m/s
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b) 5 × 10-22 kg-m/s
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c)3 × 106 kg-m/s
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d)0
Explanation
Answer: (b)
Q.30
The study of diffraction of electrons from a target, gives the wavelength associated as 0.65 Å. the energy of the electrons will be
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a) 40 eV
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b) 100 eV
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c)356 eV
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d)1000 eV
Explanation
use formula λ=√(150/V) , wavelength in ÅV=150 / (λ)2V=(12.3/0.65)2 V ≈356.0 Answer:(c)
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