# Calculate the amount of energy not used to produce electricity when a photon of wavelength 1*10^-12m hits sodium metal which has a work function of 2.3eV?

Jul 15, 2017

$E - {W}_{0} = {E}_{D}$

Hence, $1.97 \cdot {10}^{-} 13 J$

#### Explanation:

Here we're relating a few things: (1) the energy of the photon, and (2) the difference between the work function and that energy.

$E = \frac{h c}{\lambda}$
$E = \frac{\left(6.626 \cdot {10}^{-} 34 J \cdot m\right) \left(2.998 \cdot {10}^{8} \frac{m}{s}\right)}{{10}^{-} 12 m}$
$E = 1.98 \cdot {10}^{-} 13 J$

Above is the energy of the photon, now we're going to convert the work function joules, as such,

$2.3 e V \cdot \frac{1.602 \cdot {10}^{-} 19 J}{e V} \approx 3.68 \cdot {10}^{-} 19 J$

So, relating these two values to realize our differential:

$1.98 \cdot {10}^{-} 13 J - 3.68 \cdot {10}^{-} 19 J \approx 1.97 \cdot {10}^{-} 13 J$

This is fairly reasonable since a photon of that wavelength is probably gamma, ouch!