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NEET • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Mixed difficulty

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31 Medium
31 of 130
Application +4 | -1
The work function of Cesium metal is 2.15 eV. When light of frequency 6×1014 Hz6 \times {10^{14}}\,Hz is incident on metal surface, photoemission occurs. Maximum kinetic energy of emitted photon electron will be
View answer and explanation
Correct option: B
0.54×10−19 J0.54 \times {10^{ - 19}}\,J
Option B (0.54×10−19 J0.54 \times {10^{ - 19}}\,J) is correct because it is the maximum kinetic energy calculated using the photoelectric effect equation.
32 Medium
32 of 130
Remembering +4 | -1
Which is the correct ascending order of wavelengths?
View answer and explanation
Correct option: B
λgamma−ray<λX−ray<λvisible<λmicrowave{\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}
Option B (λgamma−ray<λX−ray<λvisible<λmicrowave{\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}) is correct because it represents the correct ascending order of wavelengths in the electromagnetic spectrum.
33 Easy
33 of 130
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 is correct because if the wavelengths are the same, then their momenta must be equal according to the de Broglie relation λ = h/p.
34 Medium
34 of 130
Understanding +4 | -1
A beam of light of wavelength λ\lambda and with illumination LL falls on a clean surface of sodium. If NN photoelectrons are emitted each with kinetic energy E,E, then
View answer and explanation
Correct option: B
N∝LN \propto L and E∝1λE \propto \frac{1}{\lambda }
Option B (N∝LN \propto L and E∝1λE \propto \frac{1}{\lambda }) is correct because the number of photoelectrons emitted is proportional to the light intensity, and the kinetic energy of the photoelectrons is inversely proportional to the wavelength of light.
35 Hard
35 of 130
Application +4 | -1
An electron (mass m) with an initial velocity v=v0i^v = {v_0}\hat i is in an electric field E=E0j^E = {E_0}\hat j . If initial wavelength λ0=h/mv0{\lambda _0} = h/m{v_0} , then its de Broglie wavelength at time t is given by
View answer and explanation
Correct option: C
λ01+e2E02t2m2v02\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}
Option C is correct because the de Broglie wavelength changes with the electron's velocity, which is affected by the electric field.
36 Medium
36 of 130
Understanding +4 | -1
Dual nature of radiation is shown by
View answer and explanation
Correct option: D
Photoelectric effect and diffraction
Option D (Photoelectric effect and diffraction) is correct because the photoelectric effect demonstrates the particle nature of radiation, while diffraction demonstrates the wave nature.
37 Medium
37 of 130
Analyzing +4 | -1
A photon and electron have same de-Broglie wavelength. Give that vv is the speed of electron and cc is the velocity of light. Ee,  Ep{E_e},\;{E_p} are the kinetic energy of electron and photon respectively. pe,  ph{p_e},\;{p_h} are the momentum of electron and photon respectively. Then which of the following relation is correct?
View answer and explanation
Correct option: A
EeEp=v2c\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}
Option A (Is the photon) is correct because for the same wavelength, the photon has the highest energy due to its massless nature and energy-momentum relation.
38 Medium
38 of 130
Application +4 | -1
The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3μ3\mu, the maximum velocity of the ejected photoelectrons will be
View answer and explanation
Correct option: C
More than 3v\sqrt 3 v
Option C (More than 3v\sqrt 3 v) is correct because increasing the frequency to 3n3n results in a greater increase in kinetic energy than a factor of 3, due to the work function.
39 Hard
39 of 130
Application +4 | -1
An ionisation chamber, with parallel conducting plates as anode and cathodes has 5×107  cm−35 \times {10^7}\mathrm{\;c}{\mathrm{m}^{ - 3}} electrons and the same number of singly charged positive ions per cm3\mathrm{c}{\mathrm{m}^3}. The electrons are moving toward the anode with velocity 0.4ms−1\mathrm{m}{\mathrm{s}^{ - 1}}. The current density from anode to cathode is 4μAm−2.4\mathrm{\mu A}{\mathrm{m}^{ - 2}}. The velocity of positive ions moving towards cathode is
View answer and explanation
Correct option: A
0.1  ms−10.1\;\mathrm{m}{\mathrm{s}^{ - 1}}
Option A (0.1  ms−10.1\;\mathrm{m}{\mathrm{s}^{ - 1}}) is correct because it is the velocity of the positive ions calculated using the current density and the electron velocity.
40 Medium
40 of 130
Remembering +4 | -1
The colour of the positive column in a gas discharge tube depends on
View answer and explanation
Correct option: B
The gas in the tube
Option B (The gas in the tube) is correct because the color of the light emitted from a gas discharge tube depends on the specific gas used, as different gases emit different characteristic spectra.
41 Medium
41 of 130
Understanding +4 | -1
A particle with rest mass zero is moving speed cc. The de-Broglie wavelength associated with it
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because a particle with rest mass zero moving at speed c has infinite momentum, leading to a de Broglie wavelength of zero.
42 Medium
42 of 130
Understanding +4 | -1
The linear momentum of photon is pp. The wavelength of photon is λ  ,\lambda \;, then (hh is Planck constant)
View answer and explanation
Correct option: B
λ=hp\lambda = \frac{h}{p}
Option B is correct because the wavelength of a photon is inversely proportional to its momentum, given by the equation λ = h/p.
43 Easy
43 of 130
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 because stopping potential is independent of the intensity of incident light; it depends on the frequency and the emitter material.
44 Medium
44 of 130
Remembering +4 | -1
The time taken by a photoelectron to come out after the photon strikes is approximately
View answer and explanation
Correct option: B
10−10  s{10^{ - 10}}\mathrm{\;s}
Option B (10−10  s{10^{ - 10}}\mathrm{\;s}) is correct because the photoelectric effect is nearly instantaneous.
45 Hard
45 of 130
Application +4 | -1
The maximum kinetic energy of photoelectron liberated from the surface of lithium (work function =2.35 eV=2.35 \mathrm{~eV}) by electromagnetic radiation whose electric component varies with time as E=a[1+cos⁡(2πf1t)],cos⁡(2πf2t)E=a[1+\cos (2 \pi f_1 t)], \cos \left(2 \pi f_2 t\right) where 'aa' is a constant, f1=3.6×1015 Hzf_1=3.6 \times 10^{15} \mathrm{~Hz} and f2=1.2×1015 Hzf_2=1.2 \times 10^{15} \mathrm{~Hz} is:
[Take: h=6.6×10−34 J−sh=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}]
View answer and explanation
Correct option: D
17.45 eV17.45 \mathrm{~eV}
Option D (17.45 eV17.45 \text{ eV}) is correct because it is the maximum kinetic energy calculated using the highest frequency component of the electromagnetic radiation and the work function of lithium.
46 Medium
46 of 130
Application +4 | -1
If a photon has velocity cc and frequency v,v, then which of following represents its wavelength
View answer and explanation
Correct option: A
hcE\frac{{hc}}{E}
Option A (hcE\frac{{hc}}{E}) is correct because wavelength can be expressed in terms of Planck's constant, speed of light, and energy.
47 Medium
47 of 130
Analyzing +4 | -1
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
View answer and explanation
Correct option: C
More than that of a proton
Option C (More than that of a proton) is correct because, for the same de-Broglie wavelength, the lighter particle (electron) must have a higher kinetic energy.
48 Medium
48 of 130
Application +4 | -1
A light of wavelength 4000 A∘\mathop A\limits^ \circ is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is (R=6.6×10−34(R = 6.6 \times {10^{ - 34}}Js)
View answer and explanation
Correct option: C
6.2×105  ms−16.2 \times {10^5}\;\mathrm{m}{\mathrm{s}^{ - 1}}
Option C (6.2×105 ms−16.2 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}) is correct because it is the maximum velocity calculated using the photoelectric effect equation.
49 Medium
49 of 130
Remembering +4 | -1
Momentum of a photon is p.p. The corresponding wavelength is
View answer and explanation
Correct option: D
h/ph/p
Option D is correct because the de Broglie wavelength λ is inversely proportional to the momentum p, given by the equation λ = h/p.
50 Hard
50 of 130
Application +4 | -1
An XX-ray pulse of wavelength 4.9  A∘4.9\;\mathop A\limits^ \circ is sent through a section of Wilson cloud chamber containing a super saturated gas, and tracks of photoelectron ejected from the gaseous atoms are observed. Two groups of tracks of lengths 1.40  cm1.40\;cm and 2.02  cm2.02\;cm are noted. If the range-energy relation for cloud chamber is given by R=αER = \alpha E with α=1cm/keV,\alpha = 1cm/keV, obtain the binding energies of the two levels from which electrons are emitted. Given h=6.63×10−34J−s,e=1.6×10−19Jh = 6.63 \times {10^{ - 34}}J - s,e = 1.6 \times {10^{ - 19}}J
View answer and explanation
Correct option: A
0.52  KeV0.52\;KeV
Option A (0.52  KeV0.52\;KeV) is correct because it represents the difference between the photon energy and the kinetic energy of the emitted photoelectrons, calculated using the range-energy relation.
51 Medium
51 of 130
Analyzing +4 | -1
Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is V0{V}_0 and the maximum kinetic energy of the photoelectrons is Kmax⁡{K}_{\max }. When the ultraviolet light is replaced by X{X} rays, both V0{V}_0 and Kmax⁡{K}_{\max } increases.

Statement-2: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
View answer and explanation
Correct option: C
Statement-1 is true, statement- 2 is false.
Option C is correct because Statement-1 is true (X-rays have higher energy, thus increasing stopping potential and maximum kinetic energy), but Statement-2 is false (photoelectrons have a range of speeds due to interactions within the metal, not the range of frequencies in incident light).
52 Hard
52 of 130
Understanding +4 | -1
Statement 1: Photodiodes are operated in reverse bias.

Statement 2: The current in photodiodes is proportional to light intensity.
View answer and explanation
Correct option: B
Statement 1 is true, statement 2 is true and
statement 2 is Not the correct explanation for statement
1
Option B (Statement 1 is true, statement 2 is true and statement 2 is Not the correct explanation for statement 1) is correct because photodiodes are indeed operated in reverse bias, and their current is proportional to light intensity, but the reverse bias is not *because* of the proportionality to light intensity.
53 Medium
53 of 130
Understanding +4 | -1
Einstein’s photoelectric equation states that Ek=hv−ϕ.Ek = hv - \phi . In this equation Ek{E_k} refers to
View answer and explanation
Correct option: C
Maximum kinetic energy of the emitted electrons
Option C (Maximum kinetic energy of the emitted electrons) is correct because EkE_k in Einstein's photoelectric equation represents the maximum kinetic energy of the emitted photoelectrons.
54 Hard
54 of 130
Application +4 | -1
A monochromatic light is incident on a metallic plate having work function ϕ.\phi. An electron, emitted normally to the plate from a point AA with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:

(Given: The magnitude of charge of an electron is 'e' and mass is 'm', 'h' is Planck's constant and 'c' is velocity of light. Take the magnetic field exists throughout the path of electron)
View answer and explanation
Correct option: C
8m(hcλ−ϕ)/eB\sqrt{8 \mathrm{m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}
Option C (8m(hcλ−ϕ)/eB\sqrt{8 \mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}) is correct because it represents the diameter of the circular path the electron takes in the magnetic field, derived from the kinetic energy and magnetic force equations.
55 Medium
55 of 130
Application +4 | -1
A 100 W light bulb is placed at the centre of a spherical chamber of radius 0.10m. Assume that 66% of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is
View answer and explanation
Correct option: B
1.77×10−61.77 \times {10^{ - 6}}Pa
Option B (1.77×10−61.77 \times {10^{ - 6}}Pa) is correct because it is the pressure exerted by the light calculated using the formula for radiation pressure.
56 Hard
56 of 130
Application +4 | -1
The ratio of the power of a light source S1\text{S}_1 to that the light source S2\mathrm{S}_2 is 2. S1\mathrm{S}_1 is emitting 2×10152 \times 10^{15} photons per second at 600 nm. If the wavelength of the source S2\text{S}_2 is 300 nm, then the number of photons per second emitted by S2\mathrm{S}_2 is _____ ×1014.\times 10^{14}.
View answer and explanation
Correct option: A
5
Option D (Statement-1 is false, Statement-2 is true) is correct because while the momentum is the same, energy depends on the particle's mass and velocity.
57 Medium
57 of 130
Remembering +4 | -1
Positive rays consists of
View answer and explanation
Correct option: C
Positive ions
Option C (Positive ions) is correct because positive rays are streams of positive ions.
58 Medium
58 of 130
Understanding +4 | -1
Kinetic energy of emitted cathode rays is dependent on
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because the kinetic energy of emitted cathode rays depends on both the applied voltage and the work function of the cathode material.
59 Medium
59 of 130
Remembering +4 | -1
In cathode ray oscillograph, the focusing of beam on the screen is achieved by
View answer and explanation
Correct option: C
Electric potential
Option C (Electric potential) is correct because in a cathode ray oscillograph (CRO), electric potentials are used to focus the electron beam onto the screen.
60 Easy
60 of 130
Remembering +4 | -1
A photocell works on the principle of
View answer and explanation
Correct option: B
Photoelectric effect
Option B is correct because a photocell's operation is based on the photoelectric effect, where light incident on a material causes the emission of electrons.
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