In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, then

Options

A. `E=6.8eV, lambda=6.6xx10^(-10)m`
B. `E=3.4eV, lambda=6.6xx10^(-10)m`
C. `E=3.4eV, lambda=6.6xx10^(-11)m`
D. `E=6.8eV, lambda=6.6xx10^(-11)m`

Answer

B

Explanation

The potential energy `=2xx" kinetic energy "=-2E`
`therefore" Total energy "=-2E+E=-E=-3.4eV`
`"or "E=3.4eV`.
Let p = momentum and m = mass of the electron.
`therefore" "E=(p^(2))/(2m) or p=sqrt(2mE)`
de Brogie wavelength, `lambda=(h)/(p)=(h)/(sqrt(2mE))`
On substituting the value, we get
`lambda=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx3.4xx1.6xx10^(-19)))=6.6xx10^(-10)m`