Explanation

(c)
Ionisation energy of H
$= 2.18 \times 10^{- 18}J\ atom^{- 1}$
$\therefore E_{1}$(Energy of 1st orbit of H-atom)
$= - 2.18 \times 10^{- 18}J\ atom^{- 1}$
$\therefore E_{n} = \frac{- 2.18 \times 10^{- 18}}{n^{2}}J\ atom^{- 1}$
$Z = 1\ for\ H - atom$
$\mathrm{\Delta}E = E_{4} - E_{1}$
$= \frac{- 2.18 \times 10^{- 18}}{4^{2}}—\frac{- 2.18 \times 10^{- 18}}{1^{2}}$
$= - 2.18 \times 10^{- 18} \times \left\lbrack \frac{1}{4^{2}} - \frac{1}{1^{2}} \right\rbrack$
$\mathrm{\Delta}E = hv = - 2.18 \times 10^{- 18} \times - \frac{15}{16}$
$= + 2.0437 \times 10^{- 18}J\ atom^{- 1}$
$\therefore\ \ v = \frac{\mathrm{\Delta}E}{h} = \frac{2.0437 \times 10^{- 18}J\ atom^{- 1}}{6.625 \times 10^{- 34}Js}$
$= 3.084 \times 10^{15}s^{- 1}atom^{- 1}$