The graph below is obtained for an experiment on the photoelectric effect using different frequencies of light and a given metal plate - NSC Physical Sciences - Question 10 - 2017 - Paper 1

The speed remains unchanged.

Using the equation

$E = \frac{hc}{\lambda}$

where:

• $h = 6.63 \times 10^{-34} \text{ J s}$,
• $c = 3 \times 10^{10} \text{ cm/s}$,
• $\lambda = 6 \times 10^{-7} \text{ m}$.

Substituting values:

$E_{photon} = \frac{(6.63 \times 10^{-34})(3 \times 10^{10})}{6 \times 10^{-7}} = 3.32 \times 10^{-19} \text{ J}$

This energy (3.32 x 10^-19 J) is less than the work function of the metal, therefore the photoelectric effect is NOT OBSERVED.

Using the equation:

$E = W_0 + \frac{1}{2} mv_{max}^2$

Where:

• $E = hf$, and
• $W_0 = \text{work function of the metal}$.

From the previous part, if the work function is taken from the graph at the threshold frequency ($6.8 \times 10^{14} \text{ Hz}$):

$W_0 = (6.63 \times 10^{-34})(6.8 \times 10^{14}) = 4.51 \times 10^{-19} \text{ J}$

Plugging values into the equation gives:

$6.63 \times 10^{-34} \times 7.8 \times 10^{14} = 4.51 \times 10^{-19} + \frac{1}{2} mv_{max}^2$

Thus, solving for $v_{max}$, we find that the maximum speed of the ejected photoelectron is $3.82 \times 10^{5} \text{ m/s}$.