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  4. For particles having same K.E., the de-Broglie wavelength is

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Q. For particles having same K.E., the de-Broglie wavelength is

Structure of Atom

Solution:

$\frac{\lambda_{1}}{\lambda_{2}} = \frac{m_{2}v_{2}}{m_{1}v_{1}} $
$ \frac{\frac{1}{2}m_{2}v_{2}^{2}}{\frac{1}{2}m_{1}v_{1}^{2}}\cdot\frac{v_{1}}{v_{2}} $
$ = \frac{K.E_{1}}{K.E_{2}} = \frac{v_{1}}{v_{2}}$
As $K.E_{1} = K.E_{2}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}} = \frac{v_{1}}{v_{2}}$ or $\lambda\propto v$