If the linear momentum of a proton is changed | vedclass

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

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$400 p_0$

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(C) The de Broglie wavelength $\lambda$ is related to the linear momentum $p$ by the equation $\lambda = \frac{h}{p}$.Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$.Dividing the two equations,we find the fractional change: $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.Given that the magnitude of the change in wavelength is $0.25 \%$,we have $\left| \frac{d\lambda}{\lambda} \right| = \frac{0.25}{100} = \frac{1}{400}$.Since $\left| \frac{d\lambda}{\lambda} \right| = \frac{dp}{p}$,we have $\frac{p_0}{p} = \frac{1}{400}$.Therefore,the initial linear momentum $p = 400 p_0$.

If the linear momentum of a proton is changed by $p_0$,then the de Broglie wavelength associated with the proton changes by $0.25 \%$. Then the initial linear momentum of the proton is

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