Answer
Energy required case (a) $10.2eV$ light required is ultraviolet light.
Energy required case (a) $1.889eV$ light required is visible light.
Work Step by Step
Energy difference between initial state of electron $n_{i}$ and final state of electron $n_{f}$ is given by
$\Delta E=13.6 (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})$
this suggest that energy $\Delta E$ is required to excite an electron from initial state $n_{i}$ to final state $n_{f}$.
(a) Excitation from ground state to first excited state
for ground state $n_{i}$ =1
for first excited state $n_{f}$=2
so energy required to excite electrons from ground state to first excited state is $\Delta E=13.6 (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})eV$
$\Delta E=13.6 (\frac{1}{1^{2}} - \frac{1}{n_2^{2}})eV$
$\Delta E=13.6 (1 - \frac{1}{4})eV$=$ 13.6 \times\frac{3}{4}eV$ =$10.2eV$
(b) Excitation from first excited state to second excited state
for first excited state $n_{i}$ =2
for second excited state $n_{f}$=3
so energy required to excite electrons from ground state to first excited state is $\Delta E=13.6 (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})eV$
$\Delta E=13.6 (\frac{1}{2^{2}} - \frac{1}{3_2^{2}})eV$
$\Delta E=13.6 (\frac{1}{4} - \frac{1}{9})eV$=$ 13.6 \times\frac{5}{36}eV$ =$1.889eV$
for classification of type of required light
$\lambda (in nm)=\frac{hc}{\Delta E}$= $\frac{1.24\times10^{3} eV.nm}{\Delta E (in eV)}$ (taken from equation 27.19)
for case (a) $\Delta E= 10.2eV$ so $\lambda (in nm)= \frac{1.24\times10^{3} eV.nm}{10.2 eV}$ =$121.56 nm$ this is in the range of ultraviolet radiation
for case (b) $\Delta E= 1.889eV$ so $\lambda (in nm)= \frac{1.24\times10^{3} eV.nm}{1.889 eV}$ =$656.4 nm$
this is in the range of visible light.
