NEET2020Physics-Modern Physics
NEET 2020 Physics de Broglie Wavelength MCQ Question
Type: MCQ-numerical-Medium-Class 12
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10⁻² nₘ, the potential difference is:
A
10³ V
B
10⁴ V
C
10 V
D
10² V
Correct Answer
Option B
Detailed Explanation
Using the de Broglie wavelength formula λ = h/√(2meV), solving for V gives V = 10⁴ V.
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