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Physics NEET MCQ
Quiz 1
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Q.1
in photoelectric effect work function of any metal is 2.5eV. Emitted electrons are stopped by the potential of 1.5volt then
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a)energy of incident photons is 4eV
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b) energy of incident photons is 1eV
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c)photoelectric current increases when we use photos of high frequency
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d)none of the above
Explanation
Answer: (a)
Q.2
If the wavelength of incident light changes from 4000Å to 3600Å, change in stopping potential will be
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a) +0.35V
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b) -0.35V
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c)+0.4V
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d)-0.4 V
Explanation
Answer: (a)
Q.3
The slope of the graph Kmax Vs f in photo electric effect is
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a) h
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b) h/e
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c)he
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d)e/h
Explanation
K=hf - φ on comparing with standard equation for line we get slope=hAnswer: (a)
Q.4
In photocell, energy conversion is from...
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a) chemical to electrical
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b) mechanical to electrical
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c)optical to electrical
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d)magnetic to electrical
Explanation
Answer:(c)
Q.5
The photoelectric currents at distance r1 and r2 of a light source from a photocell are I1 and I2 respectively. What is the value of I2 / I1 ?
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a) r₁ / r₂
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b) r₂ / r₁
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c) (r₁ / r₂)²
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d) (r₂ / r₁)²
Explanation
we know that current ∝ Intenisty ∝ 1/r2 Answer: (c)
Q.6
Ratio of momentum of 105eV, X-ray photon (P) with that of 105eV electron (P')
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a) P'/P=1/2
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b) P'/P=16/5
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c) P'/P=1/5
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d) P'/P=5/1
Explanation
Use formula for X-ray P=E/c=eV/c and for electrons P'=√(2eVm), here m is the mass of electron Answer: (b)
Q.7
An atom emits a photon of wavelength 1 Å. The energy of recoil of the atom will be ( mass of atom=1 amu)
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b) 1.304 eV
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d) 1.532 eV
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a) 1.304 × 10⁻²⁰ J
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c) 1.532 × 10⁻¹⁹J
Explanation
use formula Answer: (a)
Q.8
The de-Broglie wavelength of a proton and alpha particle is same, the ratio of their velocities is
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a)1:4
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b) 1:2
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c)2:1
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d)4:1
Explanation
λ=h/mv and mass of alpha particle is four times mass of protonAnswer: (d)
Q.9
The energy of an electron having de_broglie wavelength λ is
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a) h/2m
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b) h2/2mλ
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c)h2/2λ2
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d)h2/2mλ2
Explanation
Answer: (d)
Q.10
If the momentum of a particle is doubled, then its de-Broglie wavelength will become.. [ AFMC 1997]
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a)unchanged
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b) four times
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c)two times
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d)half times
Explanation
de-Broglie wave length λ=h / p If p is doubled then λ will be halved.Answer: (d)
Q.11
The threshold frequency for a photosensitive metal is 3.3×1014Hz. If light of frequency 8.2×1014Hz is incident on this metal. the cut-off voltage for the photoelectric emission is nearly [ CBSE-PMT 2011]
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a)2V
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b) 3V
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c)5V
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d)1V
Explanation
V0 is stopping potential ν0 is threshold frequencyeV0=hν -hν0 V0=h/e( ν - ν0)Answer: (a)
Q.12
Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..[ AFMC 1997]
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a) 1.16 eV
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b) 2.38 eV
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c)0.58 eV
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d)2.98 eV
Explanation
Energy of incident radiation=hc/λ Now according to Einstein's formulaK.E of photo electron=Incident energy - work functionK. E. of photo electron=2.475-1.9=0.58eVAnswer: (c)
Q.13
The work function of aluminum is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminum, then emission of electron will be ..[ AFMC 1999]
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a) depends up on the density of the surface
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b) data is incomplete
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c) not possible
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d) possible
Explanation
photoelectron will be emitted if energy of each photon is equal to or more than work function In problem energy of photon is less than work function hence emission is not possible Answer: (c)
Q.14
The de-Broglie wave length of an electron of energy 600eV is [ AFMC 2000]
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a) 4 Angstrom
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b) 2 Angstrom
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c) 1 Angstrom
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d) 0.5 Angstrom
Explanation
We can use following formula By substituting values we get Above formula can be as follows:- We know that E = p2 / (2m) p = √ (2mE) --eq(1) and λ = h/ p --eq(2) substituting value of p from equation (1) in equation (2) we get above mention formula Answer: (d)
Q.15
The magnitude of saturation photoelectric current depends upon... [ AFMC 2005]
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a) frequency
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b) intensity
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c) work function
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d) stopping potential
Explanation
The magnitude of saturation photoelectric current depends upon the intensity of radiation because higher the intensity of radiation, larger number of electrons coming out giving rise to increased intensity of photoelectric current, but frequency must be more or equal to threshold frequency Answer: (b)
Q.16
According to Einstein's photoelectric equation, the plot of kinetic energy of the emitted photoelectrons from a metal vs frequency of the incident radiation gives a straight line whose slope.. [ AFMC 2004]
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a)depends on the intensity of radiation
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b) depends on the nature of the metal used
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c)depends both on the intensity of the radiation and the metal used
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d)is the same for all metals and independent of the intensity of the radiation
Explanation
Einstein's photoelectric equation, is hν=Φ + EHere hν is energy of incident radiationΦ is work functionE is energy of electron ν=Φ/ h + (1/h) E This line is straight line ,slope of the line is 1/h which is constant So, it depends on metals usedAnswer: (d)
Q.17
For photoelectric emission, tungsten requires light of 2300Å. If light of 1800Å wave length is incident then emission ...[ AFMC 2005]
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a) takes place
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b) doesn't take place
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c)may or may not take
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d)depends on frequency
Explanation
Since light of lower wavelength than threshold wave length is used, emission will take place.Answer: (a)
Q.18
If the minimum energy of photons needed to produce photoelectric effect is 3eV, bombarding the photoelectric material by a number of photons of 2.5eV also one can get...[ AFMC 2006]
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a) photoelectrons of the same kinetic energy
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b) photoelectrons of higher kinetic energy
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c)higher current
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d)none
Explanation
As one photon is necessary to eject one photoelectron, more than one photon are not required for ejection of photoelectron. Hence, the energy of photon should be equal to or greater than the minimum energy needed to produce photoelectron.Since in the given problem, energy of photon is 2.5eV is less than the minimum energy 3eV needed to produce photoelectric effect, hence no photoelectrons will be produced. Answer:(d)
Q.19
What is incorrect about photon? ... [ AFMC 2009]
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a) Its rest mass is zero
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b) its energy is hν
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c) Its momentum is (hν)/c
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d) It does not exert pressure
Explanation
A photon exerts pressure as it has momentum and energy Answer: (d)
Q.20
Number of ejected photoelectron increases with increase ... [ CBSE-PMt 1993]
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a) in intensity of light
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b) in wave length of light
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c)in frequency of light
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d)never
Explanation
Answer: (a)
Q.21
Which of the following moving particles ( moving with same velocity) has largest wave length of matter waves [ CBSE-PMT 2002]
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a) Electron
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b) α-particle
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c)Proton
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d)Neutron
Explanation
de-Broglie wave length λ=h/ mvFor same velocity, λ ∝ 1/m Out of given particles, the mass of electron is minimum, so the associated de-Broglie wave length is maximum for electron Answer:(a)
Q.22
Which of the following statement is correct? [ CBSE-PME 1997]
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a) Photo-current increases with intensity of light
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b) Photo-current is proportional to the applied voltage
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c)Current in photocell increases with increasing frequency
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d)Stopping potential increases with increase of incident light
Explanation
Answer: (a)
Q.23
Which of the following statement is correct? [ CBSE-PME 1997]
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a) Photo-current increases with intensity of light
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b) Photo-current is proportional to the applied voltage
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c)Current in photocell increases with increasing frequency
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d)Stopping potential increases with increase of incident light
Explanation
Answer: (a)
Q.24
When photons of energy hν fall on an aluminium plate ( of work function Φ), photoelectrons of maximum kinetic energy of K are ejected. If the frequency of the radiation is doubled then ejected photoelectrons will be.. [ CBSE-PMT 2006]
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a)2K
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b)K
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c)K + hν
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d)K + Φ
Explanation
Applying Einstein's formula fro photo-electricityhν=Φ + K; here K is kinetic energy of electronIf we use 2ν frequency then let kinetic energy becomes K' soh.2ν=Φ + K'From above two equationsK'=hν + K Answer: (c)
Q.25
The X-rays cannot be diffracted by means of an ordinary grating because of .. [ CBSE-PMT 1997]
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a) high speed
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b) short wave length
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c)large wave length
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d)none of these
Explanation
We know that the X-rays are of short wave length as compared to grating constant of optical grating. As a result of this, it makes difficult to observe X-rays diffraction with ordinary gratingAnswer: (b)
Q.26
In the Davisson and Germar experiment, the velocity of electrons emitted from the electron gun can be increased by[ CBSE-PMT 2011]
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a) increasing the potential difference between the anode and filament
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b) increasing the filament current
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c) decreasing the filament current
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d) decreasing the potential difference between the anode and filament
Explanation
In Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and filament. Answer: (a)
Q.27
The number of photo electrons emitted for light of frequency ν ( higher than the threshold frequency ν0 is proportional to [ CBSE - PMT 2009]
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a) Threshold frequency ( ν0
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b) Intensity of light
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c)Frequency of light ν
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d)ν - ν0
Explanation
The number of photoelectrons emitted is proportional to the intensity of incident light. Saturation Current ∝ IntensityAnswer: (b)
Q.28
Light of two different frequencies whose photos have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5eV successively. Ratio of maximum speeds of emissions will be [ CBSE-PMT 2011]
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a)1:4
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b) 1:2
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c)1:1
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d)1:5
Explanation
The mximum kinetic energy of emitted electrons is given by E=Φ - Φ0E1=1eV - 0.5eV=0.5eVE2=2.5eV - 0.5eV=2eVNow E=½ ( m v2)Thus E ∝ v2 ∴ E1 / E2=v12 / v22 ∴ 1/4=v12 / v22 ∴ v1 / v2=1/2 Answer:(b)
Q.29
A beam of cathode rays is sunjected to crossed Electric field(E) and Magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by [ CBSE-PMT 2010]V is the potentia; difference between cathode and anode
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a)
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b)
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c)
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d)
Explanation
For no deflection of beam eE=evB ∴ v2=E2/B2 also ½ ( m v2)=eV ∴ v2=2eV / m ∴ E2/B2=2eV / m ∴ e/m=E2 / 2VB2 Answer: (d)
Q.30
The work function of a surface of a photosensitive material is 6.2 eV. The wave length of incident radiation for which the stopping potential is 5V lies in the [ CBSE-PMT 2008]
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a)Ultraviolet region
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b) Visible region
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c)Inferred region
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d)X-ray region
Explanation
Work functio Φ=6.2eVStopping potential V0=5VAs eV0=hc/ λ - Φ Or Thus the wave length of the incident radiation lies in the ultraviolet regionAnswer: (a)
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