SOLUTION —
We know that
$E_{n}=-\frac{2 \pi^{2} c^{4} m}{n^{2} h^{2}}, \quad E_{n}=-\frac{2 \cdot 179 \times 10^{-11}}{n^{2}} erg / atom$
Encrgy of the first orbit $(n=1)=\frac{-2 \cdot 179 \times 10^{-11}}{1^{2}} erg$ atom
and energy of the fourth orbit $(n=4)=\frac{-2 \cdot 179 \times 10^{-11}}{4^{2}} erg / atom$
$\Delta E=E_{4}-E_{1}=\frac{-2 \cdot 179 \times 10^{-11}}{16}+\frac{2 \cdot 179 \times 10^{-11}}{1}$
$=2 \cdot 179 \times 10^{-11}\left(1-\frac{1}{16}\right) erg / atom =\frac{2 \cdot 179 \times 10^{-11} \times 15}{16} erg$ atom
$=2 \cdot 043 \times 10^{-11} erg /$ atom .
Energy of the photon must be equal to $2 \cdot 043 \times 10^{-11} erg / atom$.
Calculation of frequency
$ E =h v$
therefore $\quad \text { frequency }(v)=\frac{E}{h} =\frac{2 \cdot 043 \times 10^{-11} erg }{6 \cdot 62 \times 10^{-27} erg . s }$
$=3 \cdot 08 \times 10^{15} s ^{-1}= 3 \cdot 0 8 \times 10^{15} H z$
Calculation of wavelength
$c=v \lambda$
$\lambda=\frac{c}{v}=\frac{3 \cdot 0 \times 10^{10} cm / s }{3 \cdot 08 \times 10^{15} s ^{-1}}= 9 \cdot 7 4 \times 10^{-4} cm$
