Question

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Q. The de-Broglie wavelength of a body moving with speed $\nu$ is $\lambda$. On its way, it losses some of its mass and gains twice the speed. Kinetic energy also increases to twice of its initial value. What will be the new value of de-Broglie wavelength?

Answer 1

Dear Student,
$$\begin{gathered} Letinitialmassofthebodyism \\ then \\ \lambda =\frac{h}{mv}------1 \\ asitskineticenergydoublesafterincreaseinvelocitytwicethen \\ letnewmassofthebodyism\text{'} \\ \frac{1}{2}m{v}^2=2\times \frac{1}{2}m\text{'}(2v{)}^2 \\ m\text{'}=\frac{m}{8} \\ sonow \\ debroglliewavelength\lambda \text{'}=\frac{h}{{\displaystyle \frac{m}{8}}\left(2v\right)}=\frac{4h}{mv}=4\lambda \\ meanswavelengthbecomes4timesofitsinitialvalue \end{gathered}$$
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