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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