The de Broglie wavelength of an electron accelerated to a potential of 400 V is approximately

Question

The de Broglie wavelength of an electron accelerated to a potential of 400 V is approximately (A) 0.03 nm (B) 0.04 nm (C) 0.12 nm (D) 0.06 nm

Tardigrade Admin · Accepted Answer

Voltage, V=400 V ∵ de-Broglie wavelength of an electron λ =(123/√V) mathringA =(12.3/√400) mathringA=(12.3/20) mathringA =0.615 mathringA ⇒ λ =0.0615 nm

Q. The de Broglie wavelength of an electron accelerated to a potential of $400 V$ is approximately

0.03 nm

0.04 nm

0.12 nm

0.06 nm

Voltage, $V = 400 V$
$∵$ de-Broglie wavelength of an electron
$λ = \frac{123}{V} \overset{˚}{A}$
$= \frac{12.3}{400} \overset{˚}{A} = \frac{12.3}{20} \overset{˚}{A}$
$= 0.615 \overset{˚}{A}$
$\Rightarrow λ = 0.0615 nm$

Q. The de Broglie wavelength of an electron accelerated to a potential of 400V is approximately