An electron is in excited state in a hydrogen-like atom. It has a total energy of `-3.4eV` .The kinetic energy of the electron is E and its de Broglie wavelength is `lambda`

Options

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

Answer

B

Explanation

`P=-2KE=-2E`
Total energy
`=PE+KE=-2E+E=-E`
`implies-E=-3.4eV`
or `E=3.4eV`
Let p is momentum and m is mass of the electron
`E=(p^(2))/(2m)`
or `p=sqrt((2mE))`
de Broglie wavelength `lamda=h/(p)=h/(sqrt((2mE)))`
`lamda=6.6xx10^(-10)m`