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

Answer

A

Explanation

`K_(n)=-E_(n)=-(-34)=3.4 eV`
`lamda=(h)/(sqrt(2mK))=(6.6xx10^(-24))/(sqrt(2xx9.1xx10^(-31)xx3.4xx1.6xx10^(-10)))`
`=6.6xx10^(-10)m`