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JEE • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Mixed difficulty

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Showing 91-120 of 316 questions

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91 Medium
91 of 316
Understanding +4 | -1
When wavelength of incident photon is decreased then
View answer and explanation
Correct option: B
Velocity of emitted photoelectron increases
Option B is correct. Decreasing the wavelength of incident photons increases their energy, which in turn increases the kinetic energy and thus the velocity of the emitted photoelectrons.
92 Medium
92 of 316
Remembering +4 | -1
Which of the following metals having highest
work function is
View answer and explanation
Correct option: D
PtPt
Option D (PtPt) is correct because Platinum (Pt) has the highest work function among the given options.
93 Medium
93 of 316
Understanding +4 | -1
The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is
View answer and explanation
Correct option: C
More than doubled
Option C (More than doubled) is correct because the kinetic energy increases by more than double due to the work function of the metal.
94 Medium
94 of 316
Application +4 | -1
A photodiode sensor is used to measure the output of a 300 W300 {~W} lamp kept 10 m10 {~m} away. The sensor has an opening of 2 cm2 {~cm} in diameter. How many photons enter the sensor if the wavelength of the light is 660 nm660 {~nm} and the exposure time is 100 ms100 {~ms}. (Assume that all the energy of the lamp is given off as light and h=6.6×10−34Js)\left.h=6.6 \times 10^{-34} {Js}\right)
View answer and explanation
Correct option: C
2.5×10132.5 \times 10^{13}
Option C (2.5×10132.5 \times 10^{13}) is correct because it is the number of photons calculated based on the power of the lamp, distance to the sensor, sensor area, wavelength, and exposure time.
95 Medium
95 of 316
Remembering +4 | -1
The wave nature of electrons was first experimentally verified independently by
View answer and explanation
Correct option: A
Davisson and Germer
The correct answer is A (Davisson and Germer) because they experimentally verified the wave nature of electrons.
96 Medium
96 of 316
Understanding +4 | -1
Which of the following transitions in a hydrogen atom emits photon of the highest frequency
View answer and explanation
Correct option: A
n = 2 to n = 1
Option A is correct because the transition from n=2 to n=1 in a hydrogen atom emits a photon with the highest frequency, corresponding to the largest energy difference.
97 Medium
97 of 316
Application +4 | -1
The energy that should be added to an electron, to reduce its de-Broglie wavelength from 10−10m{10^{ - 10}}m to 0.5×10−10m,0.5 \times {10^{ - 10}}m, will be
View answer and explanation
Correct option: B
Thrice the initial energy
The correct answer is B (Thrice the initial energy) because the energy is inversely proportional to the square of the de-Broglie wavelength, and the change in energy is calculated as E2−E1=3E1E_2 - E_1 = 3E_1.
98 Easy
98 of 316
Understanding +4 | -1
The wavelength of the waves associated with a proton and a photon are the same. Therefore, the two have equal
View answer and explanation
Correct option: C
Momentum
Option C (Momentum) is correct because, according to de Broglie's hypothesis, wavelength is inversely proportional to momentum, so equal wavelengths imply equal momenta.
99 Medium
99 of 316
Understanding +4 | -1
Given below are two statements:
Statement I: Davisson-Germer experiment establishes the wave nature of electrons.
Statement II: If electrons have wave nature, they can interfere and show diffraction.
In the light of the above statements choose the correct answer from the option given below:
View answer and explanation
Correct option: A
Both statement I and statement II are true.
Option A is correct because the Davisson-Germer experiment confirms the wave nature of electrons, and wave nature implies interference and diffraction.
100 Medium
100 of 316
Application +4 | -1
The work function for caesium is 1.8 eV1.8 \mathrm{~eV}. Light of 5000 A∘5000 ~\mathop {\text{A}}\limits^ \circ is incident on it. Calculate maximum kinetic energy of the emitted electrons. Given : h=6.6×10−34 Jsh=6.6 \times 10^{-34} \mathrm{~Js}, me=9×10−31 kg,c=3×108 ms−1m_e=9 \times 10^{-31} \mathrm{~kg}, c=3 \times 10^8 \mathrm{~ms}^{-1}.
View answer and explanation
Correct option: A
1.08×10−19 J1.08 \times 10^{-19} \mathrm{~J}
Option A (1.08×10−19 J1.08 \times 10^{-19} \mathrm{~J}) is correct because it is the maximum kinetic energy calculated using the photoelectric effect equation.
101 Medium
101 of 316
Application +4 | -1
The constant AA in the Richardson-Dushman equation for tungsten is 60×104 A m−2 K−260 \times 10^4 \mathrm{~A} \mathrm{~m}^{-2} \mathrm{~K}^{-2}. The work function of tungsten is 4.5 eV4.5 \mathrm{~eV}. A tungsten cathode having a surface area 2.0×10−5 m22.0 \times 10^{-5} \mathrm{~m}^2 is heated by a 24 W24 \mathrm{~W} electric heater. In steady state, the heat radiated by the cathode equals the energy input by the heater and the temperature becomes constant. Assuming that the cathode radiates like a blackbody, calculate the saturation current due to thermions. Take Stefan constant =6×10−8 W m−2 K−4=6 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}. Assume that the thermions take only a small fraction of the heat supplied.
View answer and explanation
Correct option: A
1.0 mA1. 0 \mathrm{~mA}
Option A (1.0 mA1. 0 \mathrm{~mA}) is correct because it is calculated using the Richardson-Dushman equation and the given parameters.
102 Medium
102 of 316
Application +4 | -1
Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 nm400 \mathrm{~nm} falls on a metal having work function 2.5 eV2.5 \mathrm{~eV}.
View answer and explanation
Correct option: A
4.2×10−25 kg m s−14.2 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
Option A (4.2×10−25kgms−14.2 \times 10^{-25} kg m s^{-1}) is correct because it accurately calculates the maximum linear momentum of the photoelectron.
103 Medium
103 of 316
Understanding +4 | -1
From the photoelectric effect experiment, following observations are made. Identify which of these are correct.
A. The stopping potential depends only on the work function of the metal.
B. The saturation current increases as the intensity of incident light increases.
C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light.
D. Photoelectric effect can be explained using wave theory of light.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: C
B only
Option C is correct because only statement B is correct: the saturation current increases as the intensity of incident light increases.
104 Medium
104 of 316
Understanding +4 | -1
A particle with rest mass m0{m_0} is moving with speed of light cc. The de-Broglie wavelength associated with it will be
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because if a particle moves at the speed of light, its momentum becomes infinite, leading to a de Broglie wavelength of zero.
105 Medium
105 of 316
Application +4 | -1
In Millikans oil drop experiment, a charged drop of mass 1.8×10−14kg1.8 \times {10^{ - 14}}kg is stationary between its plates. The distance between its plates is 0.90  cm0.90\;cm and potential difference is 2.0  kilo  volts.2.0\;kilo\;volts. The number of electrons on the drop is
View answer and explanation
Correct option: C
5
Option C (5) is correct because the number of electrons is calculated by equating the electric force to the gravitational force and solving for the number of electrons.
106 Medium
106 of 316
Application +4 | -1
In photoelectric experiment graph of stopping potential versus 1Wavelength\frac{1}{\text {Wavelength}} is drawn. Nature and reciprocal of slope of graph will be
View answer and explanation
Correct option: A
Straight line, hce\frac{h c}{e}
Option D (a/2a/\sqrt 2) is correct because at this displacement, the kinetic energy (KE) and potential energy (PE) of the particle are equal in SHM.
107 Medium
107 of 316
Analyzing +4 | -1
Assertion: Photo electric current increases if the distance between cathode and anode is increased.

Reason: Momentum of photon is directly proportional to its wavelength.
View answer and explanation
Correct option: D
If both Assertion and reason are false
Option D (If both Assertion and reason are false) is correct because increasing the distance between cathode and anode decreases photoelectric current, and the momentum of a photon is inversely proportional to its wavelength.
108 Medium
108 of 316
Application +4 | -1
A monochromatic source, emitting light of wavelength, 600 nm600 \mathrm{~nm}, has a power output of 66 W66 \mathrm{~W}. Calculate the number of photons emitted by this source in 2 minutes.
View answer and explanation
Correct option: A
2.4×1022 photons2.4 \times 10^{22} \text { photons}
Option A (2.4×1022 photons2.4 \times 10^{22} \text{ photons}) is correct because it is calculated using the energy of each photon and the total power emitted.
109 Medium
109 of 316
Application +4 | -1
The potential energy of a particle of mass mm is given by

U(x)={E0;  0≤x≤10;              x>1U\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{E_0};\;0 \le x \le 1}\\ {0;\;\;\;\;\;\;\;x > 1} \end{array}} \right.

λ1{\lambda _1} and λ2{\lambda _2} are the de-Broglie wavelengths of the particle, when 0≤x≤10 \le x \le 1 and x>1x > 1 respectively. If the total energy of particle is 2E0,2{E_0}, the ratio λ1λ2\frac{{{\lambda _1}}}{{{\lambda _2}}} will be
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because it represents the ratio of de-Broglie wavelengths calculated using the kinetic energies in the two regions.
110 Medium
110 of 316
Analyzing +4 | -1
Assertion: Work function of a metal is 8eV8 \mathrm{eV}. Two photons each having energy 5 eV\mathrm{eV} can't eject the electron from the metal.

Reason: More than one photon can't collide simultaneously with an electron.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option C is correct because the assertion is true (photons must have energy greater than the work function to eject electrons), and the reason is also true (an electron can only absorb one photon at a time).
111 Medium
111 of 316
Analyzing +4 | -1
Electric field and magnetic field in a region of space is given by E=E0j^\mathbf{E}=E_0 \hat{\mathbf{j}} and B=B0j^\mathbf{B}=B_0 \hat{\mathbf{j}}. A particle of specific charge α\alpha is released from origin with velocity v=v0i^\mathbf{v}=v_0 \hat{\mathbf{i}}. Then, path of particle
View answer and explanation
Correct option: C
is a helix with non-uniform pitch
Option C (is a helix with non-uniform pitch) is correct because the electric field causes acceleration along the y-axis, resulting in a changing velocity component parallel to the magnetic field, hence a non-uniform pitch.
112 Medium
112 of 316
Analyzing +4 | -1
Assertion: de-Broglie wavelength of an electron accelerated through a potential difference of V{V} volt is λ=12.72VA\lambda=\frac{12.72}{\sqrt{V}}{A}

Reason: de-Broglie wavelength of an electron is given by λ=hmv\lambda=\frac{h}{m v}.
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct.
The assertion is incorrect, and the reason is correct. The correct formula for de-Broglie wavelength is λ=12.27VA\lambda=\frac{{12.27}}{{\sqrt{{V}}}} {A}, not λ=12.72VA\lambda=\frac{{12.72}}{{\sqrt{{V}}}} {A}.
113 Medium
113 of 316
Application +4 | -1
A beam of 450 nm450 \mathrm{~nm} light is incident on a metal having work function 2.0 eV2.0 \mathrm{~eV} and placed in a magnetic field BB. The most energetic electrons emitted perpendicular to the field are bent in circular arcs of radius 20 cm20 \mathrm{~cm}. Find the value of BB.
View answer and explanation
Correct option: A
1.46×10−5 T1.46 \times 10^{-5} \mathrm{~T}
Option B (4.16×10−5 T4.16 \times 10^{-5} \mathrm{~T}) is correct because it is the magnetic field calculated using the kinetic energy of the emitted electrons and the radius of their circular path.
114 Medium
114 of 316
Understanding +4 | -1
Which of the following cannot be explained on the basis of photoelectric theory?

I. Instantaneous emission of photoelectrons

II. Existence of threshold frequency

III. Sufficiently intense beam of radiation can emit photoelectrons

IV. Existence of stopping potential
View answer and explanation
Correct option: C
III only
Option D (I1=I2=I3>I4{I_1} = {I_2} = {I_3} > {I_4}) is correct because the moment of inertia of the ring, disk, and cylinder are all MR²/2, while the sphere is 2/5 MR².
115 Easy
115 of 316
Understanding +4 | -1
Which of the following statement is not true about stopping potential (V0)?\left(\mathrm{V}_0\right)?
View answer and explanation
Correct option: C
It increases with increase in intensity of the incident light
Option C is correct. Stopping potential does not increase with an increase in the intensity of incident light. It depends on the frequency of incident light and the nature of the emitter material.
116 Medium
116 of 316
Understanding +4 | -1
Assertion: Speed of light is independent of frame of reference.

Reason: Speed of light is the ultimate speed of particle.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion
Option B (If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion) is correct because both statements are true based on the postulates of special relativity, but the reason doesn't explain the assertion.
117 Medium
117 of 316
Understanding +4 | -1
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal vs the frequency of the incident radiation gives a straight line whose slope
View answer and explanation
Correct option: D
is the same for all metals and independent of the intensity of the radiation
Option D (is the same for all metals and independent of the intensity of the radiation) is correct because the slope of the kinetic energy vs. frequency plot is Planck's constant, which is a universal constant.
118 Medium
118 of 316
Remembering +4 | -1
The linear momentum of a photon corresponding to wavelength of radiation λ\lambda is
View answer and explanation
Correct option: C
hλ\frac{h}{\lambda }
Option C (hλ\frac{h}{\lambda }) is correct because the momentum of a photon is given by the de Broglie relation p=hλp = \frac{h}{\lambda}.
119 Medium
119 of 316
Application +4 | -1
The minimum intensity of light to be detected by human eye is 10−10W/m2.{10^{ - 10}}W/{m^2}. The number of photons of wavelength 5.6×10−7m5.6 \times {10^{ - 7}}m entering the eye, with pupil area 10−6m2,{10^{ - 6}}{m^2}, per second for vision will be nearly
View answer and explanation
Correct option: C
300
Option C (300) is correct because it is the approximate number of photons calculated using the given intensity, area, wavelength, and Planck's constant.
120 Medium
120 of 316
Application +4 | -1
A proton and an α\mathrm{\alpha }-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength (λp/λα)\left( {{\lambda _p}/{\lambda _\alpha }} \right) is
View answer and explanation
Correct option: D
222\sqrt 2
Option D (222\sqrt{2}) is correct because the de Broglie wavelength is inversely proportional to the square root of the mass and charge when accelerated through the same potential difference.
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