$(h=6.63 \times 10^{-34} \ J \ s, c=3 \times 10^8 \ m \ s^{-1})$
\(E=\frac{h c}{\lambda} \cdot(E=h v)\)
Given, \(\lambda=45 n m=45 \times 10^{-9} m\left[\because 1 n m=10^{-9} m\right]\)
Hence, \(E=\frac{6.63 \times 10^{-34} J s \times 3 \times 10^{8} m s^{-1}}{45 \times 10^{-9} m}=4.42 \times 10^{-18} J\)
Hence, the energy corresponds to light of wavelength \(45 \;nm\) is \(4.42 \times 10^{-18} \mathrm{J}\)
$(A)$ $(a)$ $n=3,1=1, m_{l}=1, m_{s}=+\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=1, m _{s}=+\frac{1}{2}$
$(B)$ $(a)$ $n =3,1=2, m _{l}=-2, m _{s}=-\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=-1, m _{s}=-\frac{1}{2}$
$(C)$ $(a)$ $n=4,1=2, m_{l}=2, m_{s}=+\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=2, m _{s}=+\frac{1}{2}$
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