Question

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

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

Answer 1

Answer: (D)

Energy of photon corresponding to second line of Balmer series for $Li ^{2+}$ ion
$=(13.6) \times(3)^{2}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$
$=13.6 \times \frac{27}{16}$
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
$=13.6 \times \frac{9 \times 3}{16}-\frac{13.4}{4}$
$=13.6 \times\left(\frac{27-4}{16}\right)$
$\Rightarrow 19.55\, eV$