Answer
for case (a)
$f=6.9096636\times10^{5} Hz$
for case (b)
$f=8.225\times10^{5} Hz$
Work Step by Step
Energy difference between initial state of electron $n_{i}$ and final state of electron $n_{f}$ is given by
$\Delta E=13.6 (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})$ equation (1)
this suggest that energy $\Delta E$ is required to excite an electron from initial state $n_{i}$ to final state $n_{f}$.
and from equation 27.19 wavelength of light for energy difference $\Delta E$ is given by $\lambda(in nm)= \frac{hc}{\Delta E}=\frac{1.24\times10^{3}eV nm}{\Delta E (in eV)}$
since $f= \frac{c}{\lambda}$
$f=\frac{\Delta E (in eV)\times c}{1.24\times10^{3}eV.nm}$
$f=\frac{\Delta E (in eV)\times 3\times10^8m/s}{1.24\times10^{3}eV. nm}$
$f=2.41935\times10^{5}\Delta E (in eV)/s$ .....equation (2)
for case (a) $n_{i}=2 $ and $n_{f}=5 $
so from equation (1)
$\Delta E=13.6 (\frac{1}{2^{2}} - \frac{1}{5^{2}})eV$=$13.6 (\frac{1}{4} - \frac{1}{25})eV$ =$13.6\times \frac{21}{100} eV$=$2.856eV$
putting this value in equation (2)
frequency
$f=2.41935\times10^{5}\Delta E (in eV)/s$ =$2.41935\times10^{5}\times 2.856/s$
$f=6.9096636\times10^{5} Hz$
For case (b)
$n_{i}=2 $ and $n_{f}=\infty $
so from equation (1)
$\Delta E=13.6 (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})eV$=$13.6 (\frac{1}{4} - 0)eV$ =$13.6\times \frac{1}{4} eV$=$3.4eV$
putting this value in equation (2)
frequency
$f=2.41935\times10^{5}\Delta E (in eV)/s$ =$2.41935\times10^{5}\times 3.4/s$
$f=8.225\times10^{5} Hz$
