An investigation was conducted to determine the effects of changes in frequency AND intensity on the current generated in a photoelectric cell when light is incident on it - NSC Physical Sciences - Question 11 - 2016 - Paper 1

The independent variable in this experiment is the frequency of the incident light.

The threshold frequency is the minimum frequency of incident light required to eject electrons from the surface of the metal. In this case, it is given as 5,001 × 10^14 Hz.

To calculate the maximum speed of an emitted electron, we can use the energy equation:

$E_k = W_0 + hf - W_0$

Where:

• $E_k$: kinetic energy of the emitted electron
• $W_0$: work function (calculated from the threshold frequency)
• $h$: Planck's constant ($6.626 × 10^{-34} J·s$)
• $f$: frequency in experiment F ($6.50 × 10^{14} Hz$)

First, calculate the work function using the threshold frequency:

$W_0 = hf_{threshold} = 6.626 × 10^{-34} J·s × 5,001 × 10^{14} Hz = 3.313 × 10^{-19} J$

Then calculate the energy of light in experiment F:

$E = hf = 6.626 × 10^{-34} J·s × 6.50 × 10^{14} Hz = 4.303 × 10^{-19} J$

Now use the equation for kinetic energy:

$E_k = E - W_0 = 4.303 × 10^{-19} J - 3.313 × 10^{-19} J = 9.90 × 10^{-20} J$

Finally, relate kinetic energy to speed:

E_k = rac{1}{2} mv^2

where $m$ (the mass of an electron) is approximately $9.11 × 10^{-31} kg$. Solving for $v$ gives:

v = ext{sqrt}igg(rac{2E_k}{m}igg) = ext{sqrt}igg(rac{2 × 9.90 × 10^{-20} J}{9.11 × 10^{-31} kg}igg) = 1.48 × 10^5 m/s

In experiments D and E, when the intensity was doubled at the same frequency, the current increased, indicating that the number of emitted electrons increased. This suggests that the photoelectric effect is directly proportional to the intensity of incident light, as higher intensity results in more photons striking the surface, which in turn leads to more electrons being emitted.