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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 121-150 of 316 questions

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121 Medium
121 of 316
Understanding +4 | -1
The material used for making thermionic cathode must have
View answer and explanation
Correct option: B
Low work function and high melting point
Option B (Low work function and high melting point) is correct because a thermionic cathode needs a low work function for easy electron emission and a high melting point to withstand high temperatures.
122 Medium
122 of 316
Application +4 | -1
Calculate the threshold frequency of photon for photoelectric emission from a metal of work function 0.1 eV0.1 \mathrm{~eV}.
View answer and explanation
Correct option: A
2.4×1013 Hz2.4 \times 10^{13} \mathrm{~Hz}
Option A (2.4×1013 Hz2.4 \times 10^{13} \mathrm{~Hz}) is correct because the threshold frequency is calculated by dividing the work function by Planck's constant.
123 Medium
123 of 316
Analyzing +4 | -1
A moving proton and electron have the same de-Broglie wavelength. If k and pk\,\mathrm{and}\,p denote the KE and momentum, respectively. Then, choose the correct option.
View answer and explanation
Correct option: A
KP<Ke  and  Pp=Pe{K_P} < {K_e}\mathrm{ \,\,and }\,\,{P_p} = {P_e}
Option A (KP<Ke   and  Pp=Pe{K_P} < {K_e}\,\mathrm{ \,\,and }\,\,{P_p} = {P_e}) is correct because equal de-Broglie wavelengths imply equal momenta, and since the electron is lighter, it has a higher kinetic energy.
124 Medium
124 of 316
Understanding +4 | -1
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
View answer and explanation
Correct option: B
Momentum
Option B (Momentum) is correct because, according to de Broglie's hypothesis, wavelength is inversely proportional to momentum (p=h/λp = h/\lambda).
125 Medium
125 of 316
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 (λ=hp\lambda = \frac{h}{p}) is correct because the wavelength of a photon is inversely proportional to its momentum, as described by the equation λ=h/p\lambda = h/p.
126 Medium
126 of 316
Application +4 | -1
A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m then
View answer and explanation
Correct option: B
Number of electrons emitted is a quarter of the initial number
The correct answer is B (Number of electrons emitted is a quarter of the initial number) because the intensity of light, and therefore the number of emitted electrons, is inversely proportional to the square of the distance.
127 Medium
127 of 316
Analyzing +4 | -1
A photon has same wavelength as the de-Broglie wavelength of electrons. Given C=C= Speed of light V=\mathrm{V}= Speed of electron

Ee=E_e= Kinetic energy of electrons

EPh=E_{P h}= Kinetic energy of photon

Pe=P_e= Momentum of electron

PPh=P_{Ph}= Momentum of photon

Which of the following is correct?
View answer and explanation
Correct option: B
EeEPh=V2C\frac{E_e}{E_{{Ph }}}=\frac{V}{2 C}
Option B (Ee/EPh = V/2C) is correct because it correctly relates the kinetic energies of the electron and photon based on their equal wavelengths.
128 Medium
128 of 316
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, also known as canal rays, are composed of positively charged ions.
129 Medium
129 of 316
Analyzing +4 | -1
Photons absorbed in matter are converted to heat. A source emitting nn photon/s of frequency vv is used for converting of ice at 0∘C0^{\circ} {C} to water at 0∘C0^{\circ} {C}. Then, which amongst the following is incorrect regarding the time, TT taken for the conversion?
View answer and explanation
Correct option: D
Increases when the product nvn v increases
The correct answer is D (Increases when the product nvn v increases) because the time taken for conversion is inversely proportional to the product of nn and vv.
130 Medium
130 of 316
Application +4 | -1
A monochromatic source of light operating at 200 W200 \mathrm{~W} emits 4×10204 \times 10^{20} photons per second. Find the wavelength of the light.
View answer and explanation
Correct option: A
400 nm400 \mathrm{~nm}
Option A (400 nm400 \mathrm{~nm}) is correct because it is calculated using the given power and number of photons emitted per second, along with Planck's constant and the speed of light.
131 Medium
131 of 316
Understanding +4 | -1
Proton (P)(P) and electron (e)(e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, mp=1849 Me{m}_{{p}}=1849 \, {M}_{{e}} ):
View answer and explanation
Correct option: A
1:11: 1
Option A (1:11: 1) is correct because for the de-Broglie wavelength to be the same, the momentum of the proton and electron must be equal.
132 Medium
132 of 316
Understanding +4 | -1
A photon in motion has a mass equal to
View answer and explanation
Correct option: D
hv/c2hv/{c^2}
Option D (hv/c2hv/{c^2}) is correct because, according to Einstein's mass-energy equivalence, E=mc2E = mc^2, and for a photon, E=hvE = hv.
133 Medium
133 of 316
Understanding +4 | -1
The de-Broglie wavelength associated with electron in nth n^{\text {th }} Bohr orbit is (where rr is radius of nth n^{\text {th }} Bohr orbit)
View answer and explanation
Correct option: A
2πrn\frac{2 \pi r}{n}
Option A (2πrn\frac{2 \pi r}{n}) is correct because the circumference of the nth Bohr orbit is an integer multiple of the de Broglie wavelength of the electron.
134 Medium
134 of 316
Analyzing +4 | -1
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
View answer and explanation
Correct option: D
None, all have same length
Option D (None, all have same length) is correct because the de Broglie wavelength is inversely proportional to momentum, and since all rays have the same momentum, they have the same wavelength.
135 Medium
135 of 316
Understanding +4 | -1
The work function of aluminium is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminium, then emission of electron will
View answer and explanation
Correct option: C
Not possible
Option C is correct because the work function is the minimum energy required for electron emission, and the combined energy of the two photons (7.0 eV) is greater than the work function (4.2 eV).
136 Medium
136 of 316
Remembering +4 | -1
Albert Einstein was awarded Nobel Prize and his work on
View answer and explanation
Correct option: C
Photoelectric effect
Option C (Photoelectric effect) is correct because Albert Einstein was awarded the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect.
137 Medium
137 of 316
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 is correct because the kinetic energy of emitted cathode rays depends on both the applied voltage and the work function of the material.
138 Medium
138 of 316
Understanding +4 | -1
Light of intensity I1I_1 is incident on a photosensitive metal plate which gives rise to a saturation current (IS)\left(I_S\right). Now, if the intensity of the radiation is increased keeping the frequency constant, then the saturation current will be;
View answer and explanation
Correct option: B
greater than ISI_S
Option B (greater than ISI_S) is correct because, with constant frequency, increasing the intensity of light increases the number of photons, leading to more photoelectrons and a higher saturation current.
139 Medium
139 of 316
Remembering +4 | -1
The de-Broglie wavelength associated with the particle of mass mm moving with velocity vv is
View answer and explanation
Correct option: A
h/mvh/mv
Option A (h/mvh/mv) is correct because the de Broglie wavelength is inversely proportional to the momentum of the particle.
140 Medium
140 of 316
Understanding +4 | -1
The velocity of photon is proportional to (where vv is frequency)
View answer and explanation
Correct option: D
vv
Option D (vv) is correct because the velocity of a photon is directly proportional to its frequency.
141 Medium
141 of 316
Understanding +4 | -1
Assertion: A photocell is a technological application of the photoelectric effect.
Reason: Photocell is a device whose electric properties are affected by electricity.
View answer and explanation
Correct option: C
Assertion is true but Reason is false
Option C (Assertion is true but Reason is false) is correct because while a photocell does use the photoelectric effect, its electric properties are affected by *light*, not electricity.
142 Medium
142 of 316
Understanding +4 | -1
If an electron and proton are propagating in the form of waves having the same wavelength, it implies that they have the same
View answer and explanation
Correct option: A
Energy
Option A (Energy) is correct because if the wavelengths are the same, the energies are also the same according to the relation E = hc/λ.
143 Medium
143 of 316
Remembering +4 | -1
Photon of frequency vv has a momentum associated with it. If cc is the velocity of light, the momentum is
View answer and explanation
Correct option: B
hvchvc
Option D (hv/ch\mathrm{v}/c) is correct because the momentum of a photon is given by the energy divided by the speed of light, or Planck's constant times frequency divided by the speed of light.
144 Medium
144 of 316
Understanding +4 | -1
If nR{n_R} and nV{n_V} denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time, then
View answer and explanation
Correct option: B
nR>nV{n_R} > {n_V}
Option B (nR>nV{n_R} > {n_V}) is correct because red light has a longer wavelength and therefore lower energy per photon, meaning more photons are needed for the same power.
145 Medium
145 of 316
Understanding +4 | -1
The de-Broglie wavelength λ\lambda
View answer and explanation
Correct option: C
Inversely proportional to impulse
Option C (Inversely proportional to impulse) is correct because de-Broglie wavelength is inversely proportional to momentum, which is equal to impulse.
146 Hard
146 of 316
Understanding +4 | -1
In photo-electric effect, when graph is plotted for stopping potential versus frequency of incident light, for different metals, then:
View answer and explanation
Correct option: C
The slope is universal constant equal to (h/e)(h / e)
Option C is correct. The slope of the stopping potential versus frequency graph in the photoelectric effect is a universal constant equal to Planck's constant divided by the elementary charge (h/e).
147 Medium
147 of 316
Analyzing +4 | -1
Assertion: if the frequency of the incident light on a metal surface is doubled, the kinetic energy of emitted electrons is more than doubled.

Reason: The metal will provide additional energy to the emitted photoelectron for light of higher frequency than that for lower frequency.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because increasing the frequency of incident light increases the kinetic energy of emitted electrons, and the metal doesn't provide additional energy.
148 Medium
148 of 316
Understanding +4 | -1
Which of the following relations is a correct Einstein's photoelectric equation? [All symbols have their usual meaning]
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given relations are correct representations of Einstein's photoelectric equation.
149 Medium
149 of 316
Remembering +4 | -1
Positive rays were discovered by
View answer and explanation
Correct option: B
Goldstein
Option B (Goldstein) is correct because Eugen Goldstein is credited with the discovery of positive rays (also known as canal rays) in 1886.
150 Medium
150 of 316
Understanding +4 | -1
Choose the correct option from the following options given below:
View answer and explanation
Correct option: C
A classical atom based on Rutherford’s model is doomed to collapse.
Option C (A classical atom based on Rutherford’s model is doomed to collapse) is correct because, according to classical electromagnetism, the accelerating electron in Rutherford's model should radiate energy and spiral into the nucleus.
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