The de Broglie wavelength is $\lambda = \frac{h}{mv} = \frac{h}{p}$, where $h = 6.626 \times 10^{-34}$ J·s is Planck's constant and $p = mv$ is the momentum. This establishes wave-particle duality.
Back to JEE Main 2025 Modern Physics
JEE Main 2025 — Physics Modern Physics
hard
mcq
2025
Official previous-year question
Verified 30 May 2026.
Question
The de Broglie wavelength of a particle with mass $m$ moving with velocity $v$ is given by:
Options
- A
$\lambda = \frac{h}{mv}$
- B
$\lambda = \frac{mv}{h}$
- C
$\lambda = h \cdot m \cdot v$
- D
$\lambda = \frac{m}{hv}$
Solution
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Modern Physics in other years
More JEE Main Modern Physics Questions
The work function of a metal is 4.2 eV. The threshold wavelength for photoelectric emission is approximately:
2026
medium
mcq
The work function of a metal is 4.2 eV. What is the threshold wavelength for photoelectric emission? (hc = 1240 eV·nm)
2025
medium
mcq
The work function of a metal is 4.2 eV. The threshold wavelength is (hc = 1240 eV·nm):
2023
medium
mcq
In Bohrs model the ratio of the kinetic energy to the total energy of the electron in nth orbit is:
2023
hard
mcq
In hydrogen atom, the ratio of radii of first and second orbit is:
2022
easy
mcq