An electron having de- Broglie|MODERN PHYSICS|JEEMAIN 2021

JEE MAIN 2021

20-07-21 S2

Question

An electron having de- Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut – off wavelength of emitted X-ray is :

Select the correct option:

$\frac{2 m^2 c^2 \lambda^2}{h^2}$

$\frac{2 m c \lambda^2}{h}$

$\frac{\mathrm{hc}}{\mathrm{mc}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

λₑ = h / mv Kinetic energy, p² / 2m = h² / 2mλ² = hc / λ₀ λₑ = 2mc² / h

Question Tags

JEE Main

Physics

Medium

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