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)
(b)
Energy it gains $=225 \,eV$
$\lambda=\frac{h}{m v}=\frac{\lambda h}{\sqrt{2 m E}}$
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