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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 $\la

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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 $\lambda' = \lambda - \frac{0.5}{100} \lambda = 0.995 \lambda = \frac{199}{200} \lambda.$Since $\lambda = \frac{h}{p},$ the new momentum $p'$ is $p' = p + \Delta p.$Thus,$\lambda' = \frac{h}{p + \Delta p}.$Equating the two expressions for $\lambda':$$\frac{h}{p + \Delta p} = \frac{199}{200} \left( \frac{h}{p} \right).$$\Rightarrow \frac{1}{p + \Delta p} = \frac{199}{200p}.$$\Rightarrow 200p = 199(p + \Delta p).$$\Rightarrow 200p = 199p + 199 \Delta p.$$\Rightarrow 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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