de Broglie wavelength for electron: $\lambda = \frac{1.226}{\sqrt{V}}\,\text{nm}$
$$\lambda = \frac{1.226}{\sqrt{100}} = \frac{1.226}{10} = 0.123\,\text{nm}$$
Back to JEE Main 2020 Modern Physics
JEE Main 2020 — Physics Modern Physics
medium
mcq
2020
Official previous-year question
Verified 30 May 2026.
Question
The de Broglie wavelength of an electron accelerated through a potential difference of $100\,\text{V}$ is approximately:
Options
- A
$0.123\,\text{nm}$
- B
$0.0123\,\text{nm}$
- C
$1.23\,\text{nm}$
- D
$12.3\,\text{nm}$
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 de Broglie wavelength of a particle with mass m and velocity v is:
2025
hard
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
In Bohrs model the ratio of the kinetic energy to the total energy of the electron in nth orbit is:
2023
hard
mcq
The work function of a metal is 4.2 eV. The threshold wavelength is (hc = 1240 eV·nm):
2023
medium
mcq