# Question f728f

Jun 25, 2017

$2.664791 \cdot {10}^{- 20} J = 0.166549438 e V = 0.17 e V$

#### Explanation:

This will assume that either a photon has been absorbed or released by the electron. It will still work either way.

$E = h f = \frac{h c}{l} a m \mathrm{da}$, where E is the energy of the photon, h is Planck's constant (6.63x10^(-34)J s)# f is tje frequency of light, c is the speed of light (which I'll take as $3.00 x {10}^{8} m {s}^{-} 1$), and $l a m \mathrm{da}$ is the wavelength of the light.

$E = \frac{\left(6.63 \cdot {10}^{- 34}\right) \left(3.00 \cdot {10}^{8}\right)}{7.464 \cdot {10}^{- 6}} = 2.664791 \cdot {10}^{- 20} J$

If you want your answer in eV, then $\left\mid e \right\mid = 1.6 \cdot {10}^{- 19} C$, and as $V = \frac{E}{Q}$, where V is the potential difference, E is energy, and Q is $\left\mid e \right\mid$, you get $\frac{2.664791 \cdot {10}^{- 20}}{1.6 \cdot {10}^{- 19}} = 0.166549438 e V = 0.17 e V$