MCQ
A proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$
  • A
    $0.1$
  • $0.2$
  • C
    $0.3$
  • D
    $0.4$

Answer

Correct option: B.
$0.2$
b
(b)

Energy it gains $=225 \,eV$

$\lambda=\frac{h}{m v}=\frac{\lambda h}{\sqrt{2 m E}}$

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