The binding energy of an electron in the ground state of a neutral He atom is $E_0 = 24.6 ext{ eV}

Question

The binding energy of an electron in the ground state of a neutral He atom is $E_0 = 24.6 	ext{ eV}$. A single photon of energy $100 	ext{ eV}$ interacts with the atom, resulting in the ejection of both electrons. Neglecting nuclear recoil, which of the following statement(s) is/are correct? (Take $hc = 12400 	ext{ eV \AA}$)

Professor PC Thomas Classes · Answer

Energy to remove 1st $e^-$ = $24.6 	ext{ eV}$. 2nd $e^-$ is in H-like $He^+$ ($Z=2$), so $E_2 = 13.6 	imes Z^2 = 13.6 	imes 4 = 54.4 	ext{ eV}$. Total binding energy $E_{total} = 24.6 + 54.4 = 79.0 	ext{ eV}$. $KE_{max} = 100 - 79 = 21.0 	ext{ eV}$ (when the other electron emerges with zero KE). $\lambda_{max} = hc/E_{total} = 12400/79 \approx 157 	ext{ \AA}$.