Answer
(a) $\lambda = 193~nm$
(b) $P = 50~kW$
Work Step by Step
(a) We can find the energy of each photon:
$E = \frac{1.0\times 10^{-3}~J}{9.7\times 10^{14}} = 1.03\times 10^{-18}~J$
We can find the wavelength:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{1.03\times 10^{-18}~J}$
$\lambda = 193\times 10^{-9}~m$
$\lambda = 193~nm$
(b) We can find the power output during a pulse:
$P = \frac{E}{\Delta t}$
$P = \frac{1.0\times 10^{-3}~J}{20\times 10^{-9}~s}$
$P = 5.0\times 10^4~W$
$P = 50~kW$
