Physics - Modern Physics Question with Solution | TestHub

$K.E.=\frac{KZ{e}^2}{2r}$

$P.E.=\frac{-KZ{e}^2}{r}$

$T.E.=P.E.+K.E.=\frac{-KZ{e}^2}{2R}$

Therefore,$TE=-KE=\frac{PE}{2}=-3.4\text{ eV}$

So,$KE=3.14\text{ eV}$

Let$p$= momentum and$m$= mass of the electron.

$\therefore \text{E}=\frac{{\text{p}}^2}{2\text{m}}$or$\text{p}=\sqrt{2\text{mE}}$

de Broglie wavelength,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

On substituting the values, we get

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 3.4\times 1.6\times {10}^{-19}}}$

$=6.6\times {10}^{-10}\text{m}$