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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Given that the wavelength changes by $0.50\%$,the new wavelength $\lambda'$ is $\lam

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$199 \Delta p$

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Given that the wavelength changes by $0.50\%$,the new wavelength $\lambda'$ is $\lambda' = \lambda - 0.005\lambda = 0.995\lambda = \frac{199}{200}\lambda$.The new momentum $p'$ is $p + \Delta p$.Using the relation $\lambda' = \frac{h}{p'}$,we have $\frac{199}{200}\lambda = \frac{h}{p + \Delta p}$.Substituting $\lambda = \frac{h}{p}$,we get $\frac{199}{200} \cdot \frac{h}{p} = \frac{h}{p + \Delta p}$.This simplifies to $\frac{199}{200p} = \frac{1}{p + \Delta p}$.Cross-multiplying gives $199(p + \Delta p) = 200p$.$199p + 199\Delta p = 200p$.Therefore,$p = 199 \Delta p$.

If the momentum of an electron is changed by $\Delta p$,then the de-Broglie wavelength associated with it changes by $0.50\%$. The initial momentum of the electron will be

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