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

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