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) 0.03nm (B) 0.04nm (C) 0.12nm (D) 0.06nm

Tardigrade Admin · Accepted Answer

Voltage, V=400 V ∵ the de-Broglie wavelength of an electron λ =(12.3/√V) overset° A =(12.3/√400) overset° A=(12.3/20) overset° A =0.615 overset° A 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, $V = 400 V$
$∵$ the de-Broglie wavelength of an electron
$λ = \frac{12.3}{V} A^{\circ}$
$= \frac{12.3}{400} A^{\circ} = \frac{12.3}{20} A^{\circ}$
$= 0.615 A^{\circ}$
$λ = 0.0615 nm$

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

0.03nm

0.04nm

0.12nm

0.06nm