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Physics NEET MCQ
Quiz 6
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Q.1
In photoelectric effect, the photoelectric current
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a) increases when frequency of incident photons increases
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b) decreases when frequency of incident photons increases
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c)does not depend on photon frequency but only on intensity of incident beam
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d)depends both on intensity and frequency of incident beam
Explanation
Answer:(c)
Q.2
Ultra violet light of wavelength 300 nm and intensity 1.0 watt/m2 falls on the surface of a photosensitive material. If one percent of the incident photon produce photoelectron then the number of photoelectrons emitted per second from an area 1.0 cm2 of the surface is nearly [ AMU PMT 1995]
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a) 2.13 × 1011
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b) 1.51 × 1012
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c) 4.12 × 1013
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d) 9.61 ×1014
Explanation
Intenisty per sq.cm=I/104 number of incident photons falling per sq.cm As 1 percent of the incident photons produc photoelectrons, thereforenumber of photoelectrons produced per second=1013 / 6.6n=1.51 ×1012 Answer: (b)
Q.3
If we consider electrons and photons of the same wavelength, then they will have the same
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a)velocity
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b) angular momentum
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c)energy
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d)momentum
Explanation
Answer: (d)
Q.4
An electron of mass m when accelerated through a potential difference V, the de-Broglie wavelength λ. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be [ pre medical/ dental 1995]
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a) λm/M
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b) λ√(m/M)
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c)λM/m
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d)λ√(M/m)
Explanation
Answer: (b)
Q.5
Assuming photo emission to continue to take place, the factor by which the maximum velocity of the emitted photo electron changes approximately when the wavelength of the incident radiation is increased four times is [ Haryana CET 1996]
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a) 1/4
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b) 1/2
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c)2
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d)4
Explanation
Answer:(b)
Q.6
An X-ray tube operates at 10kV. The ratio of X-ray wavelength to that of de-Broglie is [ CPMT 1996]
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a) 10:1
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b) 1:10
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c) 1:100
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d) 100:1
Explanation
De-Broglie wavelength X-ray wavelength By taking ratioAnswer: (a)
Q.7
In de_broglie's equation wave-length 'λ' depends upon mass 'm' and energy 'E' according to the relation represented as [ CPMT 1996]
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a)mE1/2
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b) m-1/2E1/2
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c)m-1/2 E-1/2
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d)m1/2E-1/2
Explanation
hence option 'd' is correctAnswer: (d)
Q.8
The velocity of the most energetic electrons emitted from a metallic surface is doubled when frequency ν of incident radiation is double. The work function of this metal is [ Pb.CET 1997]
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a) zero
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b) hν/3
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c)hν/2
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d)2hν/3
Explanation
Answer: (d)
Q.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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.30
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)
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