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(C) The de Broglie wavelength $\lambda$ is related to momentum $p$ by the equation $\lambda = \frac{h}{p}$.Given that the wavelength increases by $25

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$5 p_0$

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(C) The de Broglie wavelength $\lambda$ is related to momentum $p$ by the equation $\lambda = \frac{h}{p}$.Given that the wavelength increases by $25 \%$,the new wavelength $\lambda_2 = \lambda_1 + 0.25 \lambda_1 = 1.25 \lambda_1 = \frac{5}{4} \lambda_1$.Since $\lambda \propto \frac{1}{p}$,we have $\frac{\lambda_2}{\lambda_1} = \frac{p_1}{p_2} = \frac{5}{4}$.This implies $p_2 = \frac{4}{5} p_1$.We are given that the momentum decreases by $p_0$,so $p_2 = p_1 - p_0$.Substituting the expression for $p_2$: $\frac{4}{5} p_1 = p_1 - p_0$.Rearranging the terms: $p_0 = p_1 - \frac{4}{5} p_1 = \frac{1}{5} p_1$.Therefore,the initial momentum $p_1 = 5 p_0$.

The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was

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