Question

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 least how much energy the second electron transfers to the atom in the M-shell state?

• A. $+3.4eV$
• B. $+1.51eV$
• C. $-3.4eV$
• D. $-1.51eV$

Answer 1

The correct option is B $+1.51eV$

Given that the electron is in $M$ state. This corresponds tothe principal quantum number $n=3$

From Bohr's model, energy of a state with quantum number $n$ is given by $E=-\frac{13.6}{n^2}\text{eV}$

Thus, the energy of the electron in $M$ shell is $E=-\frac{13.6}{3^2}\approx -1.51\text{eV}$

In order to ionise the atom, the minimum energy required is thus $+1.51\text{eV}$