Answer
$(a) \space v = 19.9 \space km/s$
$(b) \space v = 20.8 \space km/s$
$(c) \space v = 24.4 \space km/s$
Work Step by Step
The incident energy of the photon would get converted into Kinetic Energy.
$\therefore E_{photon} = \frac{1}{2}mv^2$ where m is mass of Hydrogen atom.
Hence, $v = \sqrt{\frac{2E_{photon}}{m}}$
$m = 1.67\times 10^{-27} kg$
$a) \space E_{photon} = 3.313\times 10^{-19}J$
$v = \sqrt{\frac{2\times 3.313\times 10^{-19}}{1.67\times 10^{-27}}} = 19.9 \space km/s$
$b) \space E_{photon} = 3.614\times 10^{-19}J$
$v = \sqrt{\frac{2\times 3.614\times 10^{-19}}{1.67\times 10^{-27}}} = 20.8 \space km/s$
$c) \space E_{photon} = 4.97\times 10^{-19}J$
$v = \sqrt{\frac{2\times 4.97\times 10^{-19}}{1.67\times 10^{-27}}} = 24.4 \space km/s$
