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
By using the relation:
$E=13.6Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$=\left(13.6\right)\times (3{)}^2[\frac{1}{2^2}-\frac{1}{4^2}]=13.6\times \frac{27}{16}$
Energy needed to eject electron from $n=2$ level in $H-$ atoms
$=13.6\times 1^2\times [\frac{1}{2^2}-\frac{1}{{\infty }^2}]\Rightarrow \frac{13.6}{4}$
$\text{Kinetic energy of ejected electron}=\text{energy difference}$
$\text{K.E. of ejected electron}=13.6\times \frac{9\times 3}{16}-\frac{13.6}{4}=13.6\times \left(\frac{27-4}{16}\right)\Rightarrow 19.55eV$