In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-Shell state. A second electron collides instantaneously with the excited hydrogen atom in the m-Shell state and ionizes it. At leas how much energy the second electron transfors to the atom is the M-shell state?

Options

A. `+34eV`
B. `+151eV`
C. `-34eV`
D. `-1.51eV`

Answer

D

Explanation

The ground state of hydrogen (n=1) is represented by K, the first excited state (n=2) is represented by L, the second excited state (n=3) is represent by M. so, the energy transferred by the second electron can be given by
`E=-13.6((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=-13.6((1)/(3^(2))-(1)/(oo^(2)))~~-1.51 eV`