Question

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. $4p_0$
• B. $\frac{p_0}{4}$
• C. $5p_0$
• D. $\frac{p_0}{5}$

Answer 1

The correct option is C $5p_0$

Initial wavelength =$\lambda$
final =$\lambda +\frac{\lambda }{4}=\frac{5\lambda }{4}$
Initial momentum = $\frac{h}{\lambda }$
final momentum = $\frac{4\lambda }{5\lambda }$
$\Delta P=\frac{h}{5\lambda }=P_0$
initial momentum =$\frac{h}{\lambda }=5P_0$