The de Broglie wavelength associated with a p | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $\lambda \propto \frac{1}{p}$.Given that the wavelength changes by $0.

Vedclass · Accepted Answer

$401\ p_0$

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $\lambda \propto \frac{1}{p}$.Given that the wavelength changes by $0.25\%$,the new wavelength $\lambda' = \lambda + \frac{0.25}{100}\lambda = 1.0025\lambda$.Since the wavelength increases,the momentum must decrease. Let the new momentum be $p' = p - p_0$.Using the relation $\lambda' = \frac{h}{p'}$,we have $1.0025\lambda = \frac{h}{p - p_0}$.Dividing this by $\lambda = \frac{h}{p}$,we get $1.0025 = \frac{p}{p - p_0}$.$1.0025(p - p_0) = p$.$1.0025p - 1.0025p_0 = p$.$0.0025p = 1.0025p_0$.$p = \frac{1.0025}{0.0025}p_0 = 401\ p_0$.

The de Broglie wavelength associated with a proton changes by $0.25\%$. If its momentum changes by $p_0$,then its initial momentum is ......

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