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: lambda = 6.6 xx 10^(-10) m`
B. `E = 3.4 eV: lambda = 6.6 xx 10^(-10) m`
C. `E = 3.4 eV: lambda = 6.6 xx 10^(-11) m`
D. `E = 6.8 eV: lambda = 6.6 xx 10^(-11) m`

Answer

B

Explanation

Potential energy `= - 2 xx` kinetic energy = `- 2 E`
Total energy `= - 2 E + E = - 3.4 eV = - E`
`E = 3.4 eV`
`P = "momentum"`, m = mass of electron
or `P = sqrt(2 m E)`
`= sqrt(2 xx 9.1 xx 10^(-31) xx 3.4 xx 1.6 xx 10^(-19)) = 10^(-24)`
de Brogile wavelength `lambda = (h)/(P) = (h)/(sqrt(2 m E)) = 6.6 xx 10^(-10) m`