Question

A proton is accelerated through $225V$. Its de Brogile wavelength is:

• A. $1.91pm$
• B. $0.2pm$
• C. $3pm$
• D. $0.4nm$

Answer 1

The correct option is A $1.91pm$

De-Brogile wavelength $\left(\lambda \right)$ of a changed particle when accelerate through a potential $v$ is given by:

$\lambda =\frac{h}{\sqrt{2mqv}}$ [$h:$ Plank's constant]

in case of proton, $m$(man of proton)=$1.67\times {10}^{-27}Kg$

$q$(charge of proton)=$1.6\times {10}^{-19}C$

Thus it gives:-$[Given,V-225v]$

$\lambda =\frac{6.62\times {10}^{-34}}{\sqrt{2\times 1.67\times {10}^{-27}\times 1.6\times {10}^{-19}\times 225}}$

$=\frac{6.62\times {10}^{-34}}{\sqrt{1.2\times {10}^{-43}}}$

$\Rightarrow \lambda \simeq 1.91\times {10}^{-12}m=1.91pm.$