Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

if e/m of electron is `1.76xx10^(11)C(kg)^(-1)` andn stopping potential is 0.71 V, then the maximum velocity of the photoelectron isA. `150kms^(-1)`B. `200kms^(-1)`C. `500kms^(-1)`D. `250kms^(-1)`

Answer» Correct Answer - C
We can write, `(1)/(2)mv^(2)=eV`
`implies v=sqrt((2eV)/(m))=sqrt(2Vxx(e)/(m))`
or `v=sqrt(2xx0.71xx1.76xx10^(11))`
`=5xx10^(5)ms^(-1)=500kms^(-1)`
Discussion
2.

The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:A. 540 nmB. 400 nmC. 310 nmD. 220 nm

Answer» Correct Answer - C
Threshold wavelength, `lamda_(0)=(hc)/(W_(0))=(12400)/(4)=3100`Å
=310 nm
Discussion
3.

A beam of cathode rays is subjected to crossed electric (E ) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given byA. `(B^(2))/(2VE^(2))`B. `(2VB^(2))/(E^(2))`C. `(2VE^(2))/(B^(2))`D. `(E^(2))/(2VB^(2))`

Answer» Correct Answer - D
As the electron bea is not deflected, then
`F_(m)=F_(e) implies Bev=Ee`
or `v=(E)/(B)` . . . (i)
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode =eV. also kinetic energy of the electron at anode `=(1)/(2)mv^(2)`. according to law of conservation of energy , `(1)/(2)mv^(2)=eV`
or `vsqrt((2eV)/(m))` . . (ii)
From eqs. (i) and (ii), we have
`sqrt((2eV)/(m))=(E)/(B)implies (e)/(m)=(E^(2))/(2VB^(2))`.
Discussion
4.

If 5% of the energy supplied to a bulb is irradiated as visible light, how many quanta are emitted per second by a 100 W lamp? (Assume, wavelength of visible light as `5.6xx10^(-5)cm`)A. `1.4xx10^(19)`B. `3xx10^(3)`C. `1.4xx10^(-19)`D. `3xx10^(4)`

Answer» Correct Answer - A
Energy radiated as visible light `=(5)/(100)xx100=5Js^(-1)`
Let n be the number of photons emitted per second.
then, `nhv=E=5`
`thereforen=(5lamda)/(hc)=(5xx5.6xx10^(-7))/((6.62x10^(-34))(3xx10^(8)))=1.4xx10^(19)`
Discussion
5.

Which of the following figure represents the variation of particle momentum and the associated de - Broglie wavelength ?A. B. C. D.

Answer» Correct Answer - B
The de-Broglie wavelength is given by
`lamda=(h)/(p)implies plamda=h`
This equationn is in the form of yx=c, which is the equation of a rectangular hyperbola. Hence, the graph given in option (b) is the correct one.
Discussion
6.

In a cathode ray oscillograph, the focusing of beam on the screen is achieved byA. convex lensesB. magnetic fieldC. electric potentialD. all of these

Answer» Correct Answer - C
In a cathode ray oscillograph, the focusing of beam on the screen is achieved by electric potential. There are two plates X and Y. X plates consists two plates `X_(1) and X_(2)` in vertical plane while Y plates also consist two plates `Y_(1) and Y_(2)` in a horizontal plane. An electric potential is applied between the X and Y plates by an external source.
Discussion
7.

When a cathode ray tube is operated at 2912 V, the velocity of electrons is `3.2xx10^(7)ms^(-1)`. Find the velocity of cathode ray if the tube is operated at 5824V.A. `2.4xx10^(7)ms^(-1)`B. `5.2xx10^(7)ms^(-1)`C. `4.525xx10^(7)ms^(-1)`D. `2.4xx10^(6)ms^(-1)`

Answer» Correct Answer - C
`v=sqrt((2eV)/(m))`
Since, e and m are constant, `(v_(1))/(v_(2))=sqrt((V_(1))/(V_(2)))`
or `v_(2)=v_(1)sqrt((V_(1))/(V_(2)))=3.2xx10^(7)sqrt((5824)/(2912))=4.525xx10^(7)ms^(-1)`
Discussion
8.

A particle of mass `1 mg` has the same wavelength as an electron moving with a velocity of `3 xx 10^(6) ms^(-1)`. The velocity of the particle isA. `3xx10^(-31)ms^(-1)`B. `2.7xx10^(-21)ms^(-1)`C. `2.7xx10^(-18)ms^(-1)`D. `9xx10^(-2)ms^(-1)`

Answer» Correct Answer - C
We know that `p=(h)/(lamda)` or `mv=(h)/(lamda)`
`therefore`As both particle and electrons having same wavelength therefore, their momentum will be equal to
`impliesm_(p)v_(p)=m_(e)v_(e)`
`implies v_(p)=(m_(e)v_(e))/(m_(p))`
`=(9.1xx10^(-31)xx3xx10^(6))/(10^(-6))`
`implies v_(p)=2.7xx10^(-18)ms^(-1)`
Discussion
9.

Monochromatic light of frequency `6.0 xx 10^(14) Hz` is produced by a laser. The power emitted is `2 xx 10^(-3)` w. The number of photons emitted, on the average, by the sources per second isA. `5xx10^(16)`B. `5xx10^(16)`C. `5xx10^(17)`D. `5xx10^(14)`

Answer» Correct Answer - A
Powerr emitted, `P=2xx10^(-3)W`
Energy of photon, `E=hv=6.6xx10^(-34)xx6xx10^(14)J`
Number of photons emitted per second
`n=(P)/(E)=(2xx10^(-3))/(6.6xx10^(-34)xx6xx10^(14))`
`=5xx10^(-15)`.
Discussion
10.

The work function for aluminium is 4.125 eV. The cut-off wavelength for photoelectric effect for aluminium will beA. 420 nmB. 350 nmC. 300 nmD. 200 nm

Answer» Correct Answer - C
The work function for aluminium,
`W_(0)=4.125eV`
`=4.125xx1.6xx10^(-19)=6.6xx10^(-19)J`
the relation for work function is given by
`W_(0)=(hc)/(lamda)` (where, `lamda` is the cut-off wavelength)
`6.6xx10^(-19)=(6.6xx10^(-34)xx3xx10^(8))/(lamda)`
`lamda=(6.6xx10^(-34)xx3xx10^(8))/(6.6xx10^(-19))`
`=(19.8xx10^(-26))/(6.6xx10^(-19))`
`=3xx10^(-7)m`
`=300xx10^(-9)m=300nm`
Discussion
11.

The cathode of a photoelectric cell is changed such that the work function changes from `(W_(1) to W_(2) (W_(2) gt W_(1))`. If the current before and after change are `I_(1)` and `I_(2)`, all other conditions remaining unchanged , then (assuming `hv gt W_(2)`)A. `l_(1)=l_(2)`B. `l_(1) lt l_(2)`C. `l_(1) gt l_(2)`D. `(W_(1))/(W_(2))=(l_(1))/(l_(2))`

Answer» Correct Answer - A
Photocurrennt depends only on intensity of light. As in given problem, only work function is changed, so current values will remain same.
Discussion
12.

From the figure describing photoelectric effect we may infer correctly that A. Na and Al bothh have the same threshold frequencyB. maximum kinetic energy for both the metals depend linearly on the frequencyC. the stopping potential are different for Na annd Al for the samme change in frequencyD. Al is a better photo sensitive material than Na

Answer» Correct Answer - B
The graph betweenn stopping potential and frequency is a straight line, so stopping potential and hence, maximum kinetic energy of photoelectrons depends linearly on the frequency.
Discussion
13.

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q isA. `1.6xx10^(-19)C`B. `3.2xx10^(-19)C`C. `4.8xx10^(-19)C`D. `8.0xx10^(-19)C`

Answer» Correct Answer - D
Balancing forces we have, `qE=mg` . . . (i)
`6pietarv=mg`
`(4)/(3)pir^(3)rhog=mg` . . . (ii)
`therefore r=((3mg)/(4pirhog))^(1//3)` . . . (iii)
Substituting the value of r in Eq. (ii), we get
`(qE)^(2)=((3)/(4pirhog))(6pietav)^(3)`
`therefore q=(1)/(E)((3)/(4pirhog))^(1//2)(6pietav)^(3//2)`
Substituting the values, we get
`q=(7)/(81pixx10^(5))sqrt((3)/(4pixx900xx9.8)xx216pi^(3))`
`xxsqrt((1.8xx10^(-5)xx2xx10^(-3))^(3))=8.0xx10^(-19)C`
Discussion