A potassium metal plate is irradiated with light of wavelength 5 x 10⁻⁷ m in an arrangement, as shown below - NSC Physical Sciences - Question 10 - 2019 - Paper 1

To calculate the energy of a single photon, we can use the equation:

E = rac{hc}{ ext{wavelength}}

Where:

• $h = 6.63 \times 10^{-34} \, \text{Js}$ (Planck's constant)
• $c = 3 \times 10^{8} \, \text{m/s}$ (speed of light)
• wavelength = $5 \times 10^{-7} \, \text{m}$

Substituting the values:

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{5 \times 10^{-7}} = 3.98 \times 10^{-19} \, \text{J}$

To prove that the ammeter will show a reading, we need to compare the photon energy with the threshold frequency. The work function can be found using:

$W_0 = h f_0$

Where $f_0$ is the threshold frequency. Substituting the values:

$W_0 = (6.63 \times 10^{-34}) (5.55 \times 10^{14}) = 3.68 \times 10^{-19} \, \text{J}$

Since $E_{photon} > W_0$ (3.98 x 10⁻¹⁹ J > 3.68 x 10⁻¹⁹ J), the photons have enough energy to eject electrons from the potassium plate, hence the ammeter will show a reading.

Increasing the intensity of the light leads to a higher number of photons striking the metal plate per second. More photons mean more electrons can be ejected, resulting in an increased current reading on the ammeter. Therefore, since more electrons are emitted per second, the ammeter reading increases.