11.1 Define the term work function in words - NSC Physical Sciences - Question 11 - 2018 - Paper 1

The metal with the higher work function is caesium. This is because caesium has a lower threshold frequency of 5.07 x 10^14 Hz compared to potassium's threshold frequency of 5.55 x 10^14 Hz. According to the photoelectric effect, a lower threshold frequency corresponds to a higher work function.

To determine if the ammeter registers a current in circuit B, we calculate the maximum energy of the incident photons using the formula:

$E = hf$

Where:

• $h = 6.63 imes 10^{-34} ext{ Js}$ (Planck's constant)
• $f = 5.55 imes 10^{14} ext{ Hz}$

Calculating:

$E = (6.63 imes 10^{-34})(5.55 imes 10^{14}) = 3.68 imes 10^{-19} ext{ J}$

The work function for potassium is given by:

$ext{Work Function} = hf_0 = 5.55 imes 10^{14} ext{ Hz} imes 6.63 imes 10^{-34} ext{ Js} = 3.68 imes 10^{-19} ext{ J}$

Since the photon energy equals the work function, the ammeter in circuit B will not register a current.

For circuit A, we will first determine the energy of the incident photons:

$E = hf = (6.63 imes 10^{-34} ext{ Js})(5.07 imes 10^{14} ext{ Hz}) = 3.36 imes 10^{-19} ext{ J}$

Now, the maximum kinetic energy (KE_max) of the ejected electrons is given by:

$KE_{max} = E - W_0$

Where:

• $W_0$ is the work function for caesium (also calculated using its threshold frequency):

$W_0 = hf_0 = 5.07 imes 10^{14} ext{ Hz} imes 6.63 imes 10^{-34} ext{ Js} = 3.36 imes 10^{-19} ext{ J}$

Thus:

$KE_{max} = 3.36 imes 10^{-19} - 3.36 imes 10^{-19} = 0 ext{ J}$

The maximum kinetic energy of an ejected electron in circuit A is therefore 0 J.

The maximum kinetic energy of the ejected electron will remain the same. Increased intensity means more photons, but not more energy per photon, thus not affecting the maximum kinetic energy.