## College Physics (4th Edition)

(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.
(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.