JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15

Download JEE MCQs for JEE Physics: Dual Nature Of Radiation and Matter

Review structured MCQ sets for JEE Physics Dual Nature Of Radiation and Matter. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Dual Nature Of Radiation and Matter

Access the complete set of multiple-choice questions for Dual Nature Of Radiation and Matter below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The radiation corresponding to the 3 to 2 transition of a hydrogen atom falls on a metal surface and produces photoelectrons. These electrons are made to enter a magnetic field of \( 3\times10^{-4} \) T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:
(a) 1.8 eV
(b) 1.1 eV
(c) 0.8 eV
(d) 1.6 eV
Show Answer & Explanation

Answer: (b) 1.1 eV

Explanation:
1. Photon energy \( =13.6\left(\frac14-\frac19\right)=1.89\ \text{eV} \).
2. Fastest electron: \( p=eBr=1.6\times10^{-19}\times3\times10^{-4}\times10^{-2}=4.8\times10^{-25} \), so \( K=\frac{p^2}{2m}=1.27\times10^{-19}\ \text{J}=0.79\ \text{eV} \).
3. Work function \( =1.89-0.79\approx1.1\ \text{eV} \)

Teacher's Note:
1. The largest radius belongs to the electrons of maximum kinetic energy.

Question 2: A certain metallic surface is illuminated with monochromatic light of wavelength \( \lambda \). The stopping potential for photoelectric current for this light is \( 3V_0 \). If the same surface is illuminated with light of wavelength \( 2\lambda \), the stopping potential is \( V_0 \). The threshold wavelength for this surface for the photoelectric effect is:
(a) \( \lambda/4 \)
(b) \( \lambda/6 \)
(c) \( 6\lambda \)
(d) \( 4\lambda \)
Show Answer & Explanation

Answer: (d) \( 4\lambda \)

Explanation:
1. \( 3eV_0=\frac{hc}\lambda-W,\quad eV_0=\frac{hc}{2\lambda}-W \); subtract: \( 2eV_0=\frac{hc}{2\lambda}\Rightarrow eV_0=\frac{hc}{4\lambda} \).
2. \( W=\frac{hc}{2\lambda}-\frac{hc}{4\lambda}=\frac{hc}{4\lambda}\Rightarrow\lambda_0=\frac{hc}{W}=4\lambda \)

Teacher's Note:
1. Eliminate the work function between the two equations to get it in terms of lambda.

Question 3:
Assertion (A): A metallic surface is irradiated by monochromatic light of frequency greater than the threshold frequency. The maximum kinetic energy and the stopping potential are \( K_{max} \) and \( V_0 \). If the frequency of the incident light is doubled, both \( K_{max} \) and \( V_0 \) are also doubled.
Reason (R): The maximum kinetic energy and stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of the incident light.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. \( K_{max}=h\nu-W \) is linear in frequency but has an intercept, so it is not proportional to it.
2. Doubling the frequency gives \( 2h\nu-W\neq2(h\nu-W) \), so A is false; R is true.

Teacher's Note:
1. A linear relation with a non-zero intercept is not a proportion.

Question 4: The surface of a metal is illuminated with light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm):
(a) 3.09 eV
(b) 1.42 eV
(c) 151 eV
(d) 1.68 eV
Show Answer & Explanation

Answer: (b) 1.42 eV

Explanation:
1. Photon energy \( =\frac{1240}{400}=3.10\ \text{eV} \).
2. \( W=3.10-1.68=1.42\ \text{eV} \)

Teacher's Note:
1. Work function equals photon energy minus the maximum kinetic energy.

Question 5: A beam of electrons (cathode rays) accelerated through a potential V is passed undeflected through crossed electric field E and magnetic field B. The specific charge of the electron is given by:
(a) \( \frac{B^2}{2VE^2} \)
(b) \( \frac{2VB^2}{E^2} \)
(c) \( \frac{2VE^2}{B^2} \)
(d) \( \frac{E^2}{2VB^2} \)
Show Answer & Explanation

Answer: (d) \( \frac{E^2}{2VB^2} \)

Explanation:
1. No deflection: \( eE=evB\Rightarrow v=\frac EB \).
2. \( eV=\frac12mv^2\Rightarrow\frac em=\frac{v^2}{2V}=\frac{E^2}{2VB^2} \)

Teacher's Note:
1. Use the velocity selector condition together with energy gained in the accelerating potential.

Question 6: Monochromatic light of frequency \( 6.0\times10^{14} \) Hz is produced by a laser. The power emitted is \( 2\times10^{-3} \) W. The number of photons emitted, on an average, by the source per second is:
(a) \( 5\times10^{16} \)
(b) \( 5\times10^{17} \)
(c) \( 5\times10^{14} \)
(d) \( 5\times10^{15} \)
Show Answer & Explanation

Answer: (d) \( 5\times10^{15} \)

Explanation:
1. \( n=\frac{P}{h\nu}=\frac{2\times10^{-3}}{6.63\times10^{-34}\times6\times10^{14}}\approx5\times10^{15} \)

Teacher's Note:
1. Photons per second equal the power divided by the energy of one photon.

Question 7: A photosensitive metallic surface has work function \( h\nu_0 \). If photons of energy \( 2h\nu_0 \) fall on this surface, the electrons come out with a maximum velocity of \( 4\times10^6 \) m/s. When the photon energy is increased to \( 5h\nu_0 \), the maximum velocity of photoelectrons will be:
(a) \( 2\times10^7 \) m/s
(b) \( 2\times10^6 \) m/s
(c) \( 8\times10^6 \) m/s
(d) \( 8\times10^5 \) m/s
Show Answer & Explanation

Answer: (c) \( 8\times10^6 \) m/s

Explanation:
1. First case: \( K_1=2h\nu_0-h\nu_0=h\nu_0 \); second: \( K_2=5h\nu_0-h\nu_0=4h\nu_0 \).
2. \( v\propto\sqrt K\Rightarrow v_2=2v_1=8\times10^6\ \text{m/s} \)

Teacher's Note:
1. Velocity scales as the square root of the maximum kinetic energy.

Question 8: In a photo-emissive cell with exciting wavelength \( \lambda \), the fastest electron has speed v. If the exciting wavelength is changed to \( 3\lambda/4 \), the speed of the fastest emitted electron will be:
(a) less than \( v\sqrt{4/3} \)
(b) \( v\sqrt{3/4} \)
(c) \( v\sqrt{4/3} \)
(d) greater than \( v\sqrt{4/3} \)
Show Answer & Explanation

Answer: (d) greater than \( v\sqrt{4/3} \)

Explanation:
1. \( \frac12mv^2=\frac{hc}\lambda-W \); \( \frac12mv'^2=\frac{4hc}{3\lambda}-W=\frac43\cdot\frac12mv^2+\frac W3 \).
2. Since W > 0, \( v'^2>\frac43v^2 \), i.e. \( v'>v\sqrt{4/3} \).

Teacher's Note:
1. The work function does not scale with the photon energy, so the kinetic energy grows by more than the factor 4/3.

Question 9: According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:
(a) Depends on the nature of the metal used
(b) Depends on the intensity of the radiation
(c) Depends both on the intensity of the radiation and the metal used
(d) Is the same for all metals and independent of the intensity of the radiation
Show Answer & Explanation

Answer: (d) Is the same for all metals and independent of the intensity of the radiation

Explanation:
1. \( K=h\nu-W \): slope = h, a universal constant.

Teacher's Note:
1. Different metals only shift the intercept.

Question 10: An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. Its final velocity will be:
(a) \( \sqrt{\frac{2eV}{m}} \)
(b) \( \sqrt{\frac{eV}{m}} \)
(c) \( \frac{eV}{2m} \)
(d) \( \frac{eV}{m} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\frac{2eV}{m}} \)

Explanation:
1. \( eV=\frac12mv^2\Rightarrow v=\sqrt{\frac{2eV}{m}} \)

Teacher's Note:
1. Work done by the field becomes kinetic energy.

Question 11: An electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength \( \lambda \). The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:
(a) \( \lambda\frac Mm \)
(b) \( \lambda\frac mM \)
(c) \( \lambda\sqrt{\frac Mm} \)
(d) \( \lambda\sqrt{\frac mM} \)
Show Answer & Explanation

Answer: (d) \( \lambda\sqrt{\frac mM} \)

Explanation:
1. \( \lambda=\frac{h}{\sqrt{2mqV}}\propto\frac1{\sqrt m} \) for equal charge and voltage.
2. \( \lambda_p=\lambda\sqrt{\frac mM} \)

Teacher's Note:
1. Same charge and voltage give the same kinetic energy, so wavelength falls with the root of mass.

Question 12: The wavelength of a 1 keV photon is \( 1.24\times10^{-9} \) m. What is the frequency of a 1 MeV photon (in Hz)?
(a) \( 1.24\times10^{15} \)
(b) \( 2.4\times10^{20} \)
(c) \( 1.24\times10^{18} \)
(d) \( 2.4\times10^{23} \)
Show Answer & Explanation

Answer: (b) \( 2.4\times10^{20} \)

Explanation:
1. \( \nu=\frac Eh=\frac{10^6\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx2.4\times10^{20}\ \text{Hz} \)

Teacher's Note:
1. Frequency is directly proportional to photon energy.

Question 13: 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 nm
(b) 400 nm
(c) 310 nm
(d) 220 nm
Show Answer & Explanation

Answer: (c) 310 nm

Explanation:
1. \( \lambda_{max}=\frac{1240}{4.0}=310\ \text{nm} \)

Teacher's Note:
1. The threshold wavelength is hc divided by the work function.

Question 14: Radiation of wavelength 180 nm ejects photoelectrons from a plate whose work function is 2.0 eV. If a uniform magnetic field of flux density \( 5.0\times10^{-5} \) T is applied parallel to the plate, what should be the radius of the path followed by photoelectrons ejected normally from the plate with maximum energy?
(a) 0.2 m
(b) 0.148 m
(c) 0.048 m
(d) 0.418 m
Show Answer & Explanation

Answer: (b) 0.148 m

Explanation:
1. Photon energy \( =\frac{1240}{180}=6.89\ \text{eV} \), so \( K=4.89\ \text{eV} \).
2. \( p=\sqrt{2mK}=1.19\times10^{-24} \), \( r=\frac{p}{eB}=\frac{1.19\times10^{-24}}{1.6\times10^{-19}\times5\times10^{-5}}\approx0.148\ \text{m} \)

Teacher's Note:
1. The electron moves perpendicular to the field, so it follows a circle of radius p/(eB).

Question 15: The de-Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it:
(a) \( \frac h{mv} \)
(b) \( hmv \)
(c) \( \frac{mh}{v} \)
(d) \( \frac m{hv} \)
Show Answer & Explanation

Answer: (a) \( \frac h{mv} \)

Explanation:
1. \( \lambda=\frac hp=\frac h{mv} \)

Teacher's Note:
1. The de Broglie relation links wavelength and momentum.

Question 16: The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in:
(a) Ultra-violet region
(b) Infra-red region
(c) Visible region
(d) X-ray region
Show Answer & Explanation

Answer: (a) Ultra-violet region

Explanation:
1. Photon energy \( =6.2+5=11.2\ \text{eV} \).
2. \( \lambda=\frac{1240}{11.2}\approx110\ \text{nm} \), which lies in the ultraviolet.

Teacher's Note:
1. Add the stopping energy to the work function to get the photon energy.

Question 17: When photons of wavelength \( \lambda_1 \) are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength \( \lambda_2 \) are used, the corresponding stopping potential is thrice the above value. If light of wavelength \( \lambda_3 \) is used, the stopping potential for this case is:
(a) \( \frac{hc}{e}\left[\frac1{\lambda_3}+\frac1{\lambda_2}-\frac1{\lambda_1}\right] \)
(b) \( \frac{hc}{e}\left[\frac1{\lambda_3}+\frac1{2\lambda_2}-\frac1{\lambda_1}\right] \)
(c) \( \frac{hc}{e}\left[\frac1{\lambda_3}-\frac1{\lambda_2}-\frac1{\lambda_1}\right] \)
(d) \( \frac{hc}{e}\left[\frac1{\lambda_3}+\frac1{2\lambda_2}-\frac3{2\lambda_1}\right] \)
Show Answer & Explanation

Answer: (d) \( \frac{hc}{e}\left[\frac1{\lambda_3}+\frac1{2\lambda_2}-\frac3{2\lambda_1}\right] \)

Explanation:
1. \( eV=\frac{hc}{\lambda_1}-W,\ 3eV=\frac{hc}{\lambda_2}-W\Rightarrow eV=\frac{hc}2\left(\frac1{\lambda_2}-\frac1{\lambda_1}\right) \)
2. \( W=\frac{hc}{\lambda_1}-eV=hc\left(\frac3{2\lambda_1}-\frac1{2\lambda_2}\right) \)
3. \( V_3=\frac{hc}{e}\left[\frac1{\lambda_3}-\frac3{2\lambda_1}+\frac1{2\lambda_2}\right] \)

Teacher's Note:
1. Find the work function from the first two stopping potentials, then apply the equation to lambda 3.

Question 18: The de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to (\( |e|=1.6\times10^{-19} \) C, \( m_e=9.1\times10^{-31} \) kg, \( h=6.6\times10^{-34} \) J s):
(a) 0.5 angstrom
(b) 1.2 angstrom
(c) 1.7 angstrom
(d) 2.4 angstrom
Show Answer & Explanation

Answer: (c) 1.7 angstrom

Explanation:
1. \( \lambda=\frac{12.27}{\sqrt{50}}\ \text{angstrom}\approx1.7\ \text{angstrom} \)

Teacher's Note:
1. Use lambda = 12.27 / sqrt(V) angstrom for electrons.

Question 19: The work function of the target material is 1.24 eV and the wavelength of the incident light is \( 4.36\times10^{-7} \) m. The retarding (stopping) potential necessary to stop the emission of photoelectrons is (in V, nearly):
(a) 1.0
(b) 1.6
(c) 2.8
(d) 4.1
Show Answer & Explanation

Answer: (b) 1.6

Explanation:
1. Photon energy \( =\frac{1240}{436}=2.84\ \text{eV} \).
2. Stopping potential \( =2.84-1.24=1.6\ \text{V} \)

Teacher's Note:
1. Stopping potential in volts equals the maximum kinetic energy in electron-volts.

Question 20: Light of two different frequencies, whose photons have energies 1 eV and 2.5 eV, successively illuminates a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons is 1 : n. The value of n is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Kinetic energies: \( 1-0.5=0.5\ \text{eV} \) and \( 2.5-0.5=2\ \text{eV} \).
2. \( \frac{v_1}{v_2}=\sqrt{\frac{0.5}{2}}=\frac12 \), so n = 2.

Teacher's Note:
1. Speed varies as the square root of kinetic energy.

Question 21: If the de-Broglie wavelength of an electron accelerated through a potential difference of V volt is \( \sqrt{\frac nV} \) angstrom, then the value of n is (nearest):
(a) 100
(b) 150
(c) 200
(d) 250
Show Answer & Explanation

Answer: (b) 150

Explanation:
1. \( \lambda=\frac{h}{\sqrt{2meV}}=\frac{12.27}{\sqrt V}\ \text{angstrom} \), so \( n=12.27^2\approx150 \)

Teacher's Note:
1. Square the numerical constant 12.27.

Free study material for Physics

Practice MCQs for JEE Physics Dual Nature Of Radiation and Matter

Download Multiple Choice Questions: Dual Nature Of Radiation and Matter (JEE Physics)

Review structured objective questions for JEE Physics Dual Nature Of Radiation and Matter. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Dual Nature Of Radiation and Matter

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Physics to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for JEE Physics

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15?

You can get most exhaustive JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Dual Nature Of Radiation and Matter MCQs Set 15 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.