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(D) The de Broglie wavelength $\lambda$ is related to momentum $P$ by the equation $\lambda = \frac{h}{P}$.Taking the logarithmic derivative,we get $\

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$200\,\Delta P$

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(D) The de Broglie wavelength $\lambda$ is related to momentum $P$ by the equation $\lambda = \frac{h}{P}$.Taking the logarithmic derivative,we get $\frac{d\lambda}{\lambda} = -\frac{dP}{P}$.For small changes,we can write $\frac{|\Delta \lambda|}{\lambda} = \frac{\Delta P}{P}$.Given that the wavelength changes by $0.5\%$,we have $\frac{\Delta \lambda}{\lambda} = 0.5\% = \frac{0.5}{100} = \frac{1}{200}$.Substituting this into the relation,we get $\frac{\Delta P}{P} = \frac{1}{200}$.Therefore,the initial momentum $P = 200\,\Delta P$.

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

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