A photon of energy $6.40 \times 10^{-19}$ J is incident on a metal plate - Scottish Highers Physics - Question 16 - 2019

The kinetic energy of the photoelectron is also related to its speed by the equation:

$E_{kinetic} = \frac{1}{2} mv^2$

Where:

• $m$ is the mass of the electron, approximately $9.11 \times 10^{-31} \text{ kg}$.

Rearranging for speed ($v$) gives:

$v = \sqrt{\frac{2E_{kinetic}}{m}}$

Substituting the known values:

$v = \sqrt{\frac{2 \times 2.20 \times 10^{-19}}{9.11 \times 10^{-31}}}$

Calculating this yields:

$v \approx 6.95 \times 10^5 \text{ ms}^{-1}$

Thus, the maximum speed of the photoelectron is approximately $6.95 \times 10^5 \text{ ms}^{-1}$.