2023 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (3) $3hf_0$

Given information

What we know:

The threshold frequency of the metal is $f_0$.

The light that falls on the metal has frequency $f = 4f_0$.

What the question asks

We want: the maximum kinetic energy $K_{\max}$ of the ejected electrons.

What each letter means:

$f$ ($\text{Hz}$): the frequency of the light that falls on the metal.

$f_0$ ($\text{Hz}$): the threshold frequency of the metal.

$h$ ($\text{J s}$): Planck's constant.

$\phi$ ($\text{J}$): the work function of the metal.

$K_{\max}$ ($\text{J}$): the maximum kinetic energy of an ejected electron.

Relevant theory

The energy required to get an electron free

from the metal surface is known as the work function of that metal. If it is denoted by the

symbol ϕ, the maximum kinetic energy (K max) that an emitted electron can have, should

be given by Kmax = hf - ϕ.

Since the work function is a constant depending on the metal, when the frequency of the

incident ray on the metal surface is reduced gradually, the kinetic energy received by the

election decreases gradually and at a certain frequency it becomes zero. If the frequency

is further reduced beyond this, as the energy given by the photon is not suffi cient to emit

an electron, the photoelectric eff ect does not take place. Therefore, the frequency at

which the kinetic energy becomes zero, is the threshold frequency.

The first quote gives the energy that is left for the electron. The second quote says that at the threshold frequency this energy is zero.

So at the threshold frequency the whole light energy $hf_0$ is spent on getting the electron free. That means $\phi = hf_0$.

For example, think of paying at the gate of a fair. Each coin is one packet of light energy. The gate fee is the work function.

A coin that is just worth the fee gets you in with nothing left over. A bigger coin gets you in and leaves change.

The change is the kinetic energy of the electron.

Required equations

$K_{\max} = hf - \phi$

$0 = hf_0 - \phi$

$\phi = hf_0$

$K_{\max} = hf - hf_0$

Applying it to this question

The energy comes out as a multiple of $hf_0$, so no change of units is needed. We keep $h$ and $f_0$ as symbols.

$hf_0$

$\phi = hf_0$

$4hf_0$

$hf = h \times 4f_0$

$3hf_0$

$K_{\max} = 4hf_0 - hf_0$

Reasoning and calculation

The light packet carries $4hf_0$.

Freeing the electron costs $hf_0$.

The rest, $4hf_0 - hf_0 = 3hf_0$, becomes the kinetic energy. That is option (3).

Checks

$a - b$

The unit: $h$ is in $\text{J s}$ and $f_0$ is in $\text{Hz} = \text{s}^{-1}$. So $hf_0$ is in $\text{J}$, an energy, as it should be.

Does it make sense?

Put the answer back: $K_{\max} + \phi = 3hf_0 + hf_0 = 4hf_0 = h \times 4f_0$. That is the light we were given.

The kinetic energy must be less than the light energy $4hf_0$, because some energy was spent. $3hf_0$ is less, so it fits.

Conclusion

$K_{\max} = 3hf_0$

The maximum kinetic energy is $K_{\max} = hf - hf_0 = 4hf_0 - hf_0 = 3hf_0$.

The answer is (3).

Take the work function from the threshold frequency: $\phi = hf_0$.

Then take it away from the light energy: $K_{\max} = hf - hf_0$. Check by putting it back: $K_{\max} + hf_0$ should give $hf$.

Exam tip