If the momentum of a proton is changed by p_0 | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$,which implies $\frac{d\l

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$400 \, p_0$

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$,which implies $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.For small changes,we can write $\left| \frac{\Delta \lambda}{\lambda} \right| = \left| \frac{\Delta p}{p} \right|$.Given $\left| \frac{\Delta \lambda}{\lambda} \right| = 0.25 \% = \frac{0.25}{100} = \frac{1}{400}$ and $\Delta p = p_0$.Substituting these values,we have $\frac{p_0}{p} = \frac{1}{400}$.Therefore,the initial momentum $p = 400 \, p_0$.

If the momentum of a proton is changed by $p_0$,the corresponding change in the de Broglie wavelength is $0.25 \%$. The initial momentum of the proton is:

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