In the photoelectric effect electrons are emitted from the surface of a metal when the incoming light of intensity I has a frequency f that exceeds a certain value f₀, the threshold frequency - Leaving Cert Physics - Question c - 2021

In this scenario, when the frequency f is still greater than the threshold frequency f₀, and the intensity I increases (while keeping f constant), the number of emitted electrons remains the same. However, the emitted electrons will have greater energy and speed due to the increased intensity which corresponds to more energy in total being transferred to the electrons.

If the incoming frequency f is less than the threshold frequency f₀, even with an increase in intensity I while keeping f constant, no electrons will be emitted. This is because the energy of the incoming photons is insufficient to overcome the work function of the metal.

To find the threshold frequency f₀, we use the formula: $f₀ = \frac{W}{h}$ where W is the work function and h is Planck's constant. We can express the work function in joules (1 eV = 1.6 x 10⁻¹⁹ J):

W = 2.6 eV = 2.6 × 1.6 × 10⁻¹⁹ J = 4.16 × 10⁻¹⁹ J.

Using h = 6.63 × 10⁻³⁴ J·s: $f₀ = \frac{4.16 × 10^{-19}}{6.63 × 10^{-34}} = 6.28 × 10^{14} Hz$

To find the maximum speed of the emitted electrons, we can use the conservation of energy. The energy of the emitted electrons can be given by: $E = hf - W$ Substituting for W, and knowing h and f from the previous calculations, we have: $vm = \frac{E}{\frac{1}{2} mv^2}$. Using the values from above, we can solve for v:

1. Convert wavelength to frequency: $f = \frac{c}{\lambda} = \frac{3 × 10^8 m/s}{440 × 10^{-9} m} = 6.82 × 10^{14} Hz$
2. Calculate E: $E = hf = (6.63 × 10^{-34})(6.82 × 10^{14}) = 4.52 × 10^{-19} J$
3. Use kinetic energy formula to find maximum speed v: $\frac{1}{2} mv^2 = E - W = E - 4.16 × 10^{-19}$
4. Solving for v, gives: v = 2.8 × 10^6 m/s.