Question

The binding energy of an electron in the ground state of He is equal to $24.6 eV$. Calculate the energy required to remove both the electrons

Answer 1

SOLUTION : The cnergy expression is $E_{n}=-\frac{13 \cdot 6}{n^{2}} Z^{2} eV$

Here (a) $He \rightarrow He ^{+}+e$

(b) $He ^{+} \rightarrow He ^{2+}+e$

(Here $z=2$ for He)

for equation (b) $E_{2}=-\frac{13 \cdot 6}{1^{2}} \cdot 2^{2}=-54 \cdot 4 eV$

for equation (a) $E_{1}=-24 \cdot 6 eV$

Thus to remove both the electron from $H e$. Binding energy $=24 \cdot 6+54 \cdot 4=79 \cdot 0 eV$