Question

The energy of an electron in the ground state (n = 1) for \(He ^+\) ion is –x J, then that for an electron in n = 2 state for \(Be^{3+}\) ion in J is 

(1) -x 

(2) \(-\frac {x}{9}\)

(3) -4x

(4) \(-\frac {4}{9}x\)

Answer 1

Correct option is (1) -x 

\(E_n=-R_H\left(\frac{z^2}{n^2}\right) J\)

For \(\mathrm{He}^{+}(n=1),\)

\(E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4 R_H \)

\(\therefore\) \(R_H = \frac {x}{4}\)

For \(Be ^{3+}\) (n = 2),

\(E_n = -R_H \left(\frac {Z^2}{n^2}\right)J\)

\(=-\frac{x}{4} \times\left(\frac{4 \times 4}{2 \times 2}\right)=-x J\)