8.1 Photons and the photoelectric effect — Practise

Work through these on your own. Change nanometres to metres and electronvolts to joules before you substitute, and write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A photon has a frequency of 5.0e14 Hz. Find its energy.
Find the energy of a photon of wavelength 400 nm, in joules and in electronvolts.
A photon has an energy of 2.0 eV. Find its frequency.
Find the wavelength of a photon of energy 3.0e-19 J.
A metal has a work function of 2.30 eV. Find its threshold wavelength.
Light of wavelength 300 nm falls on a metal of work function 2.00 eV. Find the maximum kinetic energy of the emitted electrons, in eV.
Light of wavelength 250 nm falls on a metal of work function 2.00 eV. Find the stopping potential.
Explain why increasing the intensity of light below the threshold frequency releases no photoelectrons, however bright the light is.
Explain why the maximum kinetic energy of the photoelectrons does not depend on the intensity of the light.
A metal releases electrons only when the light has a wavelength shorter than 620 nm. Find its work function in eV.
A laser of wavelength 633 nm emits light with a power of 2.0 mW. Find the number of photons it emits each second.
In an experiment the photon energy is 3.00 eV with an uncertainty of 0.5%, and the work function is 2.00 eV with an uncertainty of 1.0%. Find the maximum kinetic energy and its absolute uncertainty.

Answers and concise solutions

1. The photon energy

Answer: $E = hf = 6.63 \times 10^{-34} \times 5.0 \times 10^{14} = 3.3 \times 10^{-19}\,\text{J}$ .

2. The photon energy

Answer: $E = \dfrac{hc}{\lambda} = 4.97 \times 10^{-19}\,\text{J} = 3.11\,\text{eV}$ .

3. The frequency

Answer: $E = 2.0 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19}\,\text{J}$ , so $f = \dfrac{E}{h} = \dfrac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.8 \times 10^{14}\,\text{Hz}$ .

4. The wavelength

Answer: $\lambda = \dfrac{hc}{E} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.0 \times 10^{-19}} = 6.6 \times 10^{-7}\,\text{m}$ , or 660 nm.

5. The threshold wavelength

Answer: $\phi = 2.30 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}\,\text{J}$ , so $\lambda_0 = \dfrac{hc}{\phi} = 5.4 \times 10^{-7}\,\text{m}$ , or 540 nm.

6. The maximum kinetic energy

Answer: The photon energy is $\dfrac{hc}{\lambda} = 6.63 \times 10^{-19}\,\text{J} = 4.14\,\text{eV}$ . Then $E_{k(\text{max})} = 4.14 - 2.00 = 2.14\,\text{eV}$ .

7. The stopping potential

Answer: The photon energy is 4.97 eV, so $E_{k(\text{max})} = 2.97\,\text{eV}$ . Then $V_s = \dfrac{E_{k(\text{max})}}{e} = 2.97\,\text{V}$ .

8. Explain why increasing the intensity of light below the threshold frequency releases no photoelectrons, however bright the light is.

Each photon carries an energy $hf$ , which is fixed by the frequency.
One electron needs at least the work function to escape.
One photon is absorbed by one electron, so the electron cannot combine the energy of two photons.
A brighter beam has more photons, but each one still carries too little energy, so no electrons escape.

9. Explain why the maximum kinetic energy of the photoelectrons does not depend on the intensity of the light.

A photoelectron is released when it absorbs one photon.
The energy of a photon is $hf$ , which depends only on the frequency, not on the intensity.
The work function is a property of the metal and does not change with the light.
So the maximum kinetic energy, $hf - \phi$ , is also fixed. A brighter beam releases more electrons, not faster ones.

10. The work function

Answer: $\phi = \dfrac{hc}{\lambda_0} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{620 \times 10^{-9}} = 3.2 \times 10^{-19}\,\text{J} = 2.0\,\text{eV}$ .

11. The number of photons each second

Answer: Each photon carries $\dfrac{hc}{\lambda} = 3.14 \times 10^{-19}\,\text{J}$ . The number each second is $\dfrac{2.0 \times 10^{-3}}{3.14 \times 10^{-19}} = 6.4 \times 10^{15}$ .

12. The kinetic energy and its uncertainty

Answer: The kinetic energy is $3.00 - 2.00 = 1.00\,\text{eV}$ . For a difference the absolute uncertainties add: $0.015 + 0.020 = 0.035\,\text{eV}$ . So the answer is $1.00 \pm 0.035\,\text{eV}$ , a percentage uncertainty of 3.5%.