#### Answer

(a) A single ultraviolet photon has greater energy as its wavelength is shorter.
(b) The energy of an infrared photon is $9.94\times 10^{-20}~J$
The energy of an ultraviolet photon is $2.84\times 10^{-18}~J$
(c) $2.0\times 10^{21}$ infrared photons are emitted each second.
$7.0\times 10^{19}$ ultraviolet photons are emitted each second.

#### Work Step by Step

(a) $E = \frac{hc}{\lambda}$
A photon with a longer wavelength has less energy. Therefore, a single ultraviolet photon has greater energy as its wavelength is shorter.
(b) We can find the energy of an infrared photon:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{2.0\times 10^{-6}~m}$
$E = 9.94\times 10^{-20}~J$
The energy of an infrared photon is $9.94\times 10^{-20}~J$
We can find the energy of an ultraviolet photon:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{7.0\times 10^{-8}~m}$
$E = 2.84\times 10^{-18}~J$
The energy of an ultraviolet photon is $2.84\times 10^{-18}~J$
(c) We can find the number of infrared photons emitted each second:
$\frac{200~W}{9.94\times 10^{-20}~J} = 2.0\times 10^{21}~s^{-1}$
$2.0\times 10^{21}$ infrared photons are emitted each second.
We can find the number of ultraviolet photons emitted each second:
$\frac{200~W}{2.84\times 10^{-18}~J} = 7.0\times 10^{19}~s^{-1}$
$7.0\times 10^{19}$ ultraviolet photons are emitted each second.