# Question #0a47a

Jan 18, 2018

It can be done, but it’s a multi-stage calculation.

#### Explanation:

Firstly find the momentum from the wavelength (remembering to convert to m), next get a kinetic energy and finally convert to eV.

1) $p = \frac{h}{\lambda} = \frac{6.63 \times {10}^{-} 34}{1 \times {10}^{-} 10} = 6.63 \times {10}^{-} 24$ kgm/s

2) ${E}_{k} = {p}^{2} / \left(2 m\right) = {\left(6.63 \times {10}^{-} 24\right)}^{2} / \left(2 \times 1.675 \times {10}^{-} 27\right) = 1.31 \times {10}^{-} 20$ J

3) $e V = \frac{J}{1.6} \times {10}^{-} 19 = \frac{1.31 \times {10}^{-} 20}{1.6 \times {10}^{-} 19} = 0.08$ eV

Check the sums, but the logic is good!