Define the term work function of a metal. One of the metals in the table below was used in this experiment. Use a suitable calculation to identify the metal used. Calculate X.

10.1 Define the term work function of a metal.

The work function ($\phi$) of a metal is the minimum amount of energy required to eject an electron from the surface of that metal.

10.2 Use a suitable calculation to identify the metal used.

Step 1: Identify the work function from the graph. The graph shows $v^2_{max}$ versus the energy of incident photons. According to the photoelectric effect equation, E_k_{max} = E_{photon} - \phi. When $v^2_{max} = 0$, the kinetic energy is zero, meaning $E_{photon} = \phi$. This point is the x-intercept of the graph. From the graph, the line intersects the x-axis at $4 \times 10^{-19}$ J. Therefore, the work function of the metal is $\phi = 4 \times 10^{-19}$ J.

Step 2: Compare the work function to the given table.

• Barium: $4 \times 10^{-19}$ J
• Chromium: $5 \times 10^{-19}$ J
• Aluminium: $7 \times 10^{-19}$ J

The metal used in the experiment is Barium.

10.3 Calculate X.

Step 1: State the photoelectric effect equation relating kinetic energy to photon energy and work function. The maximum kinetic energy of an ejected electron is given by: $E_{k_{max}} = E_{photon} - \phi$ We also know that $E_{k_{max}} = \frac{1}{2}mv^2_{max}$. Substituting this into the equation: $\frac{1}{2}mv^2_{max} = E_{photon} - \phi$ Rearranging for $v^2_{max}$: $v^2_{max} = \frac{2}{m}(E_{photon} - \phi)$ Where $m$ is the mass of an electron ($9.11 \times 10^{-31}$ kg) and $\phi$ is the work function ($4 \times 10^{-19}$ J from 10.2).

Step 2: Substitute the given values to calculate X. X represents $v^2_{max}$ when $E_{photon} = 5.4 \times 10^{-19}$ J. $X = \frac{2}{9.11 \times 10^{-31} \text{ kg}}(5.4 \times 10^{-19} \text{ J} - 4 \times 10^{-19} \text{ J})$ $X = \frac{2}{9.11 \times 10^{-31}}(1.4 \times 10^{-19})$ $X = (2.19539 \times 10^{30}) \times (1.4 \times 10^{-19})$ $X = 3.0735 \times 10^{11} \text{ m}^2 \cdot \text{s}^{-2}$ The y-axis label is $v^2_{max} (\times 10^{11} \text{ m}^2 \cdot \text{s}^{-2})$, so X is the numerical coefficient. $X \approx \boxed{\text{3.07}}$

10.4 How will the maximum velocities of the ejected electrons change when the intensity of the incident light is increased? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer.

The maximum velocities of the ejected electrons REMAINS THE SAME.

Reason: The maximum kinetic energy (and thus maximum velocity) of photoelectrons depends only on the frequency (or energy) of the incident photons and the work function of the metal. Increasing the intensity of the incident light only increases the number of photons striking the surface per unit time, which in turn increases the number of photoelectrons emitted, but it does not change the energy of individual photons or the maximum kinetic energy of each ejected electron.