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(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Taking the derivati

Vedclass · Accepted Answer

$400\ P_o$

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(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Taking the derivative,we have $\frac{d\lambda}{dp} = -\frac{h}{p^2} = -\frac{\lambda}{p}$.For small changes,we can write $\frac{\Delta \lambda}{\lambda} = -\frac{\Delta p}{p}$.Taking the magnitude,we have $\left| \frac{\Delta \lambda}{\lambda} \right| = \left| \frac{\Delta p}{p} \right|$.Given that the change in momentum $\Delta p = P_o$ and the percentage change in wavelength $\frac{\Delta \lambda}{\lambda} = 0.25\% = \frac{0.25}{100} = \frac{1}{400}$.Substituting these values into the equation: $\frac{P_o}{p} = \frac{1}{400}$.Therefore,the initial momentum $p = 400\ P_o$.

If a change in the momentum of an electron by $P_o$ results in a change of $0.25\%$ in its de-Broglie wavelength,then the initial momentum is:

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