Question

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

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

Answer 1

The correct option is D $19.55eV$

The energy of a proton that corresponds to the second line of the Balmer series for $L{i}^{2+}$ ion is:
$=\left(13.6\right)\times (3{)}^2[\frac{1}{2^2}-\frac{1}{4^2}]=13.6\times \frac{27}{16}$
$\therefore$ $Z$ for $Li=3$
$n_1=2$ (Balmer series)
$n_2=4$ (Balmer series second line)
Energy needed to eject electron from n = 2 level in H-atom:
$=13.6\times 1^2\times [\frac{1}{2^2}-\frac{1}{\infty }]=\frac{13.6}{4}$
$\text{Since}Z=1(H-atom)$
$n_1=2$ (first excited state), $n_2=\infty$
K.E. of ejected electron will be the difference in these two energies:
$=13.6\times \frac{27}{16}-\frac{13.6}{4}$
$=13.6\left(\frac{27-4}{16}\right)=19.55eV$