91Ó°ÊÓ

Using the formula, \(\Delta E = k{z^2}\left( {\frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}}} \right)\)

Where, E is the energy of photon, z is the atomic number, \({n_1}\,and\,\,{n_2}\) are the initial and final levels in which the electron transition takes place.

\(\begin{aligned}\Delta E = k{z^2}\left( {\frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}}} \right)\\ = 2.179 \times {10^{ - 18}} \times 4 \times \left( {\frac{1}{{25}} - \frac{1}{9}} \right)\\ = 6.102 \times {10^{ - 19}}\,J\end{aligned}\)

\(\)Hence, the energy of the photon is\({\bf{6}}{\bf{.102 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}\,{\bf{J}}\).

\(\begin{aligned}\lambda = \frac{{hc}}{E}\\ = \frac{{6.626 \times {{10}^{ - 34}}m{s^{ - 1}} \times 2.998 \times {{10}^8}m{s^{ - 1}}}}{{6.102 \times {{10}^{ - 19}}\,J}}\\ = 3.25 \times {10^{ - 7}}\,m\end{aligned}\)

Hence, the wavelength is\({\bf{3}}{\bf{.25 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}\,{\bf{m}}\).