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(C) The de-Broglie wavelength $\lambda$ is related to momentum $p$ by the equation $\lambda = \frac{h}{p}$.Taking the derivative,we get $d\lambda = -\

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$400 p_0$

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(C) The de-Broglie wavelength $\lambda$ is related to momentum $p$ by the equation $\lambda = \frac{h}{p}$.Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$.Dividing by $\lambda$,we have $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.Taking the magnitude,$\left| \frac{\Delta \lambda}{\lambda} \right| = \left| \frac{\Delta p}{p} \right|$.Given that the change in momentum $\Delta p = p_0$ and the percentage change in wavelength $\frac{\Delta \lambda}{\lambda} = 0.25\% = \frac{0.25}{100} = \frac{1}{400}$.Substituting these values,we get $\frac{p_0}{p} = \frac{1}{400}$.Therefore,the original momentum $p = 400 p_0$.

When the momentum of a proton is changed by an amount $P_0$,the corresponding change in the de-Broglie wavelength is found to be $0.25\%$. Then,the original momentum of the proton was

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