Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2025 · 2 Apr · Shift 2 · Q60
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2025 · 2 Apr · Shift 2 · Q60

Dual Nature of Radiation question

2025 · 2 Apr · Shift 2 · Q60

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron with mass ' m ' with an initial velocity (t=0)v→=v0i^(v0>0)(\mathrm{t}=0) \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0\gt 0\right)(t=0)v=v0​i^(v0​>0) enters a magnetic field B→=B0j^\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{j}B=B0​j^​. If the initial de-Broglie wavelength at t=0\mathrm{t}=0t=0 is λ0\lambda_0λ0​ then its value after time ' t ' would be :
  1. A
    λ01−e2 B02t2 m2\frac{\lambda_0}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}1− m2e2 B02​t2​​λ0​​
  2. B
    λ0\lambda_0λ0​
  3. C
    λ01+e2 B02t2 m2\lambda_0 \sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}λ0​1+ m2e2 B02​t2​​
  4. D
    λ01+e2 B02t2 m2\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}1+ m2e2 B02​t2​​λ0​​
View written solutionFree

Correct answer: B

  1. Given data
  • Initial velocity: v⃗(0)=v0i^\vec v(0)=v_0\hat iv(0)=v0​i^
  • Magnetic field: B⃗=B0j^\vec B=B_0\hat jB=B0​j^​
  • Initial de-Broglie wavelength: λ0\lambda_0λ0​

We need the de-Broglie wavelength after time ttt.

  1. Magnetic force on the electron

The Lorentz force is

F⃗=q(v⃗×B⃗)\vec F = q(\vec v\times \vec B)F=q(v×B)

For an electron, q=−eq=-eq=−e.

Since v⃗⊥B⃗\vec v \perp \vec Bv⊥B, the magnetic force is always perpendicular to the velocity.

Therefore, the magnetic field can only change the direction of velocity, not its magnitude.

So,

∣v⃗∣=v0for all time|\vec v| = v_0 \quad \text{for all time}∣v∣=v0​for all time

Hence momentum magnitude remains constant:

p=m∣v⃗∣=mv0p = m|\vec v| = mv_0p=m∣v∣=mv0​
  1. de-Broglie wavelength

The de-Broglie wavelength is

λ=hp\lambda = \frac{h}{p}λ=ph​

Since p=mv0p=mv_0p=mv0​ remains constant,

λ=hmv0=λ0\lambda = \frac{h}{mv_0} = \lambda_0λ=mv0​h​=λ0​

for all time ttt.

  1. Check options
  • A: λ01−e2B02t2m2\dfrac{\lambda_0}{\sqrt{1-\frac{e^2B_0^2t^2}{m^2}}}1−m2e2B02​t2​​λ0​​ — incorrect, wavelength should not change.
  • B: λ0\lambda_0λ0​ — correct.
  • C: λ01+e2B02t2m2\lambda_0\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}λ0​1+m2e2B02​t2​​ — incorrect.
  • D: λ01+e2B02t2m2\dfrac{\lambda_0}{\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}}1+m2e2B02​t2​​λ0​​ — incorrect.
  1. Final answer

The de-Broglie wavelength after time ttt remains unchanged:

λ0\boxed{\lambda_0}λ0​​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

PreviousNext

More from Dual Nature of Radiation

  • The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is2025 · MCQ
  • The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is2025 · MCQ
  • A small mirror of mass m is suspended by a massless thread of length l. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c= speed of light in…2025 · MCQ
  • Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R…2025 · MCQ
  • A photoemissive substance is illuminated with a radiation of wavelength λi​ so that it releases electrons with de-Broglie wavelength λe​. The longest wavelength of radiation that can emit photoelectron is λo​.…2025 · MCQ
  • An electron is released from rest near an infinite non-conducting sheet of uniform charge density '−σ'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical…2025 · Numerical
  • The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of2025 · MCQ
  • An electron in the ground state of the hydrogen atom has the orbital radius of 5.3×10−11 m while that for the electron in third excited state is 8.48×10−10 m. The ratio of the de Broglie…2025 · MCQ