Question

If radiation corresponding to second line of "Balmer series" of $L{i}^{2+}$ ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:

• A. $2.55eV$
• B. $4.25eV$
• C. $11.25eV$
• D. $19.55eV$

Answer 1

The correct option is D $19.55eV$

Energy of photon corresponding to second line of Balmer series for $L{i}^{2+}$ ion
$=\left(13.6\right)\times (3{)}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$=13.6\times \frac{27}{16}$
Energy needed to eject electron from $n=2$ level in H-atom;
$=13.6\times 1^2\times \left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]\Rightarrow \frac{13.6}{4}$
K.E of ejected electron $=$ energy difference
$=13.6\times \frac{9\times 3}{16}-\frac{13.4}{4}=13.6\times \left(\frac{27-4}{16}\right)$
$\Rightarrow 19.55eV$