End of chapter exercises

We first calculate the energy of the photons:

\begin{align*} E & = \frac{hc}{\lambda}\\ & = \frac{(\text{3} \times \text{10}^{\text{8}})(\text{6,63} \times \text{10}^{-\text{34}})}{\text{400} \times \text{10}^{-\text{9}}}\\ & = \text{2,01} \times \text{10}^{-\text{19}}\text{ J} \end{align*}

Next convert the work function energy into J:

\begin{align*} \text{2,9} \times \text{1,6} \times \text{10}^{-\text{19}} = \text{4,64} \times \text{10}^{-\text{19}}\text{ J} \end{align*}

The energy of the photons is less than the work function of calcium and so no electrons will be emitted.

\begin{align*} E & = \frac{hc}{\lambda}\\ & = \frac{(\text{3} \times \text{10}^{\text{8}})(\text{6,63} \times \text{10}^{-\text{34}})}{\text{990} \times \text{10}^{-\text{9}}}\\ & = \text{4,97} \times \text{10}^{-\text{19}}\text{ J} \end{align*}