NEET2019Physics-Modern Physics
NEET 2019 Physics de Broglie Wavelength MCQ Question
Type: MCQ-numerical-Medium-Class 12
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) (mₑ = 9 × 10⁻³¹ kg)
A
12.2 × 10⁻¹⁴ m
B
12.2 nₘ
C
12.2 × 10⁻¹³ m
D
12.2 × 10⁻¹² m
Correct Answer
Option D
Detailed Explanation
The de Broglie wavelength is given by λ = h/p, where p is the momentum. The momentum can be calculated using the kinetic energy gained by the electron, eV = 0.5 * m * v². Solving for λ gives approximately 12.2 × 10⁻¹⁴ m.
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