Answer
$a) 3.18\times 10^{-18} Joule $ ; UV Region
$b) 7.82\times 10^{-20} Joule $ ; IR Region
$c) 3.60 \times 10^{-21} Joule $; IR Region
Work Step by Step
Energy of a Photon, $E = h\nu$
where $h = Planck Constant = 6.626 \times 10^{-34} Joule-sec$
& $\nu = frequency$
a) Here, $\nu = 4.80 \times 10^{15} S^{-1} $
Hence, $E = 6.626\times 10^{-34}\times 4.80 \times 10^{15} Joule = 3.18 \times 10^{-18} $Joule per photon
Also, frequency $\nu = \frac{c}{\lambda} $
where c = speed of light in vacuum = $3 \times 10^{8} m/s$
and $\lambda $ = wavelength of the radiation
Hence, $\lambda = \frac{c}{\nu}$
Here, $\lambda = \frac{3\times 10^{8}}{4.80\times 10^{15}}$ = $6.25\times 10^{-8} m$ = 62.5 nm . This wavelength of radiation lies in the UV region.
b) $\nu = 1.18 \times 10^{14} s^{-1}$
$\therefore E = 6.626\times 10^{-34}\times 1.18 \times 10^{14} = 7.82 \times 10^{-20} Joule $
and $\lambda = \frac{3\times 10^{8}}{1.18\times 10^{14}} = 2.54 \times 10^{-6}m = 2.54 \mu m$. This wavelength of radiation lies in IR region.
c) b) $\nu = 5.44 \times 10^{12} s^{-1}$
$\therefore E = 6.626\times 10^{-34}\times 5.44 \times 10^{12} Joule = 3.6 \times 10^{-21} J$
and $\lambda = \frac{3\times 10^{8}}{5.44\times 10^{12}} = 5.51 \times 10^{-5}m = 55.1 \mu m$. This wavelength of radiation also lies in IR region.
