An experiment is conducted to investigate the relationship between the frequency of light incident on a metal and the maximum kinetic energy of the emitted electrons from the surface of the metal - NSC Physical Sciences - Question 10 - 2020 - Paper 1

The physical quantity represented by X on the graph is the frequency of the incident light, measured in hertz (Hz).

The metal that needs incident light with the largest wavelength for the emission of electrons is potassium. This is because potassium has the lowest work function, which corresponds to a higher threshold wavelength.

The term work function refers to the minimum energy required for an electron to be emitted from the surface of a metal. It represents the energy barrier that must be overcome for the electron to escape the metal surface.

To calculate the work function (Φ) of platinum, we can use the formula:

$E_k = h f - ext{Φ}$

In this case, we rearrange this to:

$ext{Φ} = h f - E_k$

Using given values, where $E_k = 6.63 \times 10^{-34} J s \times 1.5 \times 10^{15} Hz$:

Calculating:

$E_k = 9.945 \times 10^{-19} J$

Then:

$ext{Φ}_{ ext{platinum}} = E_k$

Thus, the work function is approximately $9.945 \times 10^{-19} J$.

To find the frequency ($f$) associated with the maximum kinetic energy of the emitted electrons, we use the equation:

$E_k = \frac{1}{2} mv^2$

Substituting the given maximum velocity:

Where $m$ is the mass of the electron ($9.11 \times 10^{-31} kg$), we first calculate the kinetic energy:

$E_k = \frac{1}{2} (9.11 \times 10^{-31} kg) (5.60 \times 10^5 m/s)^2 = 1.44 \times 10^{-19} J$

Using the work function to find the frequency:

$E_k = h f - \text{Φ}$

Solving for $f$:

$f = \frac{E_k + \text{Φ}}{h}$

After substituting the values of Φ ($9.945 \times 10^{-19} J$) and $h (6.63 \times 10^{-34} J s)$, we can calculate:

Thus, the resulting frequency is approximately $2.41 \times 10^{15} Hz$.