Answer
a)$3.18\times10^{-18}Joule$ ; UV Region
b)$7.82\times10^{-20}Joule$ ; IR Region
c)$3.60\times10^{-21}Joule$; IR Region
Work Step by Step
Energy of a Photon, $E=h\nu$
where $h=Planck Constant=6.626\times10^{-34}Joule.sec$
& $\nu$=frequency
a) Here, $\nu=4.80\times10^{15}S^{-1}$
Hence, $E=6.626\times10^{-34}\times4.80\times10^{15}Joule=3.18\times10^{-18}Joule per photon$
Also, frequency $\nu = \frac{c}{\lambda}$
where c = speed of light in vacuum = $3\times10^{8}m/s$
and $\lambda$ = wavelength of the radiation
Hence, $\lambda = \frac{c}{\nu}$
Here, $\lambda= \frac{3\times10^{8}}{4.80\times10^{15}} = 6.25\times10^{-8}m = 62.5 nm $. This wavelength of radiation lies in the UV region.
b) $\nu=1.18\times10^{14}S^{-1}$
$\therefore E=6.626\times10^{-34}\times1.18\times10^{14}=7.82\times10^{-20}Joule$
and $\lambda=\frac{3\times10^{8}}{1.18\times10^{14}} = 2.54\times10^{-6}m = 2.54\mu m.$ This wavelength of radiation lies in IR region.
c) $\nu=5.44\times10^{12}S^{-1}$
$\therefore E=6.626\times10^{-34}\times5.44\times10^{12}Joule=3.6\times10^{-21}J$
and $\lambda=\frac{3\times10^{8}}{5.44\times10^{12}}=5.51\times10^{-5}m=55.1\mu m.$ This wavelength of radiation also lies in IR region.
