Answer
$(a) \space v = 17.3 \space km/s$
$(b) \space v = 631 \space km/s$
$(c) \space v = 77.4 \space km/s$
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 Helium atom($^4He$).
Hence, $v = \sqrt{\frac{2E_{photon}}{m}}$
$m = 4.0026m_u =4\times 1.66 \times 10^{-27} = 6.6465 \times 10^{-27} kg$
$a) \space E_{photon} = 9.939\times 10^{-19}J$
$v = \sqrt{\frac{2\times 9.939\times 10^{-19}}{6.6465\times 10^{-27}}} = 17.3 \space km/s$
$b) \space E_{photon} = 1.325\times 10^{-15}J$
$v = \sqrt{\frac{2\times 1.325\times 10^{-15}}{6.6465\times 10^{-27}}} = 631 \space km/s$
$c) \space E_{photon} = 1.99\times 10^{-23}J$
$v = \sqrt{\frac{2\times 1.99\times 10^{-23}}{6.6465\times 10^{-27}}} = 77.4 \space km/s$
