The photoelectric effect

Show that each X‑ray photon produces about 2000 light photons.

The photocathode has a work function of 2.3 eV.

Explain what is meant by the work function.

Calculate the maximum kinetic energy of the electrons leaving the photocathode.

12 X‑ray photons are detected every minute.

Use your answer to (a) to calculate the current I leaving the photocathode. Assume that all the photons of light produce photoelectrons.

State one other assumption you have made to enable you to calculate the current I in (b)(iii).

Calculate the work function of potassium in eV.

This question is about a photoelectron emitted perpendicular to plate E and with an initial kinetic energy of 4.8 × 10^–20 J.

Show that the magnitude of deceleration of this photoelectron is 8.8 × 10¹² m s^–2. (i)

Show that the initial speed of the photoelectron is about 3 × 10⁵ m s^–1. (ii)

Calculate the time t taken by the photoelectron to travel from plate E to plate C.

Using the axes shown below, sketch a graph of kinetic energy E_k against distance x from plate E.

Explain, in terms of photons, what happens to the ammeter reading when light of frequency greater than 6.3 × 10¹⁴ Hz is now incident on plate E.

Define the work function of a metal.

The work function of potassium is 2.3 eV.

Potassium emits electrons from its surface when blue light is incident on it. Extremely intense red light produces no electrons.

Explain these observations in terms of photons and their energy.

Light from a laser is incident on some potassium in a vacuum. Electrons are emitted. The wavelength of the light is 320 nm.

Calculate the shortest de Broglie wavelength of the emitted electrons.

Electromagnetic radiation is incident on a metal of work function 2.3 eV. The maximum kinetic energy (KE) of the photoelectrons is 1.7 eV.

The frequency of this incident electromagnetic radiation is kept the same but its intensity is doubled.

What is the maximum KE of the photoelectrons now?

A 1.7 eV

B 2.9 eV

C 3.4 eV

D 4.0 eV

Your answer

Use Fig. 19.2 to show that the p.d. across the 50 resistor is 0.50 V.

Calculate the e.m.f. E of the power supply.

The LED emits blue light of wavelength 4.7 × 10^–7 m.

Estimate the number of blue light photons emitted from the LED per second.

number of photons per second =........ s^–1

The light from the LED is incident on a metal of work function 2.3 eV.

Explain, with the help of a calculation, whether or not photoelectrons will be emitted from the surface of the metal.

Draw the straight line of best fit on Fig 20.1 and determine the gradient of the line.

gradient =........ V m

Calculate the percentage difference between your value in (ii) and the accepted value of the Planck constant.

Identify the two types of errors shown by the data in Fig. 20.1 and suggest how you could have refined the experiment to reduce or eliminate these errors.

Fig. 20.2 shows a gold-leaf electroscope with a clean zinc plate.

The zinc plate, metal stem and the gold-leaf are given a negative charge by briefly connecting the zinc plate to the negative electrode of a high-voltage supply.

The gold leaf is fully diverged.

The position of the leaf is not affected by intense white light from a table lamp incident on the zinc plate. The gold leaf collapses very quickly when low-intensity ultraviolet radiation from a mercury lamp is incident on the zinc plate.

Explain these observations in terms of photons.

Fig. 6.1 shows a single photomultiplier tube and its internal components. The tube can detect gamma photons in high-energy physics experiments. A single gamma photon incident on the scintillator crystal generates many photons of blue light. These visible light photons travel to the photocathode where they are converted into photoelectrons. The number of electrons is then multiplied in the photomultiplier tube with the help of electrodes called dynodes. A short pulse of electric current is produced at the output end of the photomultiplier tube.

Fig. 6.2 shows a section through the scintillator crystal in air.

The refractive index of the scintillator crystal for visible light is 1.69. The refractive index of air is 1.00. Calculate the critical angle C for this crystal.

Explain why the visible light inside the scintillator crystal follows the path shown in Fig. 6.2.

A high energy gamma photon passing through the scintillator crystal converts some of its energy into visible light photons of mean wavelength 450 nm.

Show that the energy of a single photon of wavelength 450 nm is less than 3 eV.

The photocathode is coated with potassium which has a work function of 2.3 eV. Each emitted photoelectron is accelerated by a potential difference of 100 V between the photocathode and a metal plate, called the first dynode.

Show that the maximum kinetic energy of an emitted electron at the photocathode is very small compared to its kinetic energy of 100 eV at the first dynode.

2000 photoelectrons are released from the photocathode. Each photoelectron has enough energy to release four electrons from the first dynode at the collision. These four electrons are then accelerated to the next dynode where the process is repeated. There are 9 dynodes in the photomultiplier tube. The total number of electrons collected at the anode for each photoelectron is 4⁹.

The pulse of electrons at the anode lasts for a time of 2.5 × 10⁻⁹ s.

Calculate the average current due to this pulse.

Electromagnetic radiation is incident on a negatively charged zinc plate. Electrons are emitted from the surface of the plate when a weak intensity ultraviolet source is used. Electrons are not emitted at all when an intense visible light from a lamp is used.

Explain these observations.

The maximum wavelength of the electromagnetic radiation incident on the surface of a metal which causes electrons to be emitted is 2.9 10⁻⁷ m.

Calculate the maximum kinetic energy of electrons emitted from the surface of the metal when each incident photon has energy of 5.1 eV.