MCQ
The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$, if its momentum is decreased by $p_0$. The initial momentum was
- A$4 p_0$
- B$\frac{p_0}{4}$
- ✓$5 p_0$
- D$\frac{p_0}{5}$
✓
Answer
Correct option: C.
$5 p_0$
c
(c)
(c)
$\frac{\lambda_2-\lambda_1}{\lambda_1}=\frac{25}{100}$
$\frac{\lambda_2}{\lambda_1}=\frac{5}{4} \Rightarrow \frac{P_1}{P_2}=\frac{5}{4}$
$P_2-P_1=-P_0$
$\frac{4 P_1}{5}-P_1=-P_0$
$P_1=5 P_0$
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