Question

Questions

12. The diagram shows a zinc plate exposed to weak ultraviolet (UV)
light. The UV light causes electrons to be emitted from the
surface of the plate.
13.
14.
UV light
-
a) Name this phenomenon.
b)
Initially, the plate is neutral in charge. State and explain
the effect on the charge of the plate as the zinc plate is
exposed to the UV light.
c)
State and explain the effect on the rate of emission of
electrons when the intensity of the UV light is increased.
d)
In a data book, the work function energy of zinc is quoted as 4.24 eV. Explain what is meant
by the work function energy (no calculations are necessary).
electrons
a) The concept of the photon was important in the
development of physics throughout the last century.
Explain what is meant by a photon.
b) The diagram shows a photocell. When the metal surface
is exposed to electromagnetic radiation, photoelectrons
are ejected. The collector collects the photoelectrons
and the sensitive ammeter indicates the presence of a
tiny current.
i. For a certain frequency and intensity of
radiation, the ammeter shows a current of
1.2 x 10-7 A. Calculate:
a) Einstein's photoelectric equation may be written as: hf = 0 + mvmax².
Identify the terms hf, and mvmax².
b) The surface of sodium metal is exposed to electromagnetic radiation of wavelength
6.5 x 10-7 m. This wavelength is the maximum for which photoelectrons are released.
i. Calculate the threshold frequency.
zinc plate
ii. Show that the work function energy of the metal is 1.9 eV.
c) For a particular wavelength of incident light, sodium releases photoelectrons. State how the
rate of releases of photoelectrons changes with the intensity of light is doubled. Explain
your answer.
glass bulb
metal.
1. The energy of each photon.
2. The maximum kinetic energy of each photoelectron.
3. The current in the photocell.
radiation
vacuum
1. The charge reaching the collector in 5.0 s.
2. The number of photoelectrons reaching the collector in 5.0 s.
ii. The work function energy of the metal is 3.5 x 10-19 J and the incident radiation has
a frequency of 7.0 x 10¹4 Hz. Calculate the maximum kinetic energy of an ejected
Mes photoelectron.
iii. The intensity of the incident radiation is doubled, but the wavelength is kept
constant. State the effect this has on each of the following:

Answer 1

Photoelectric effect and photons

a) The phenomenon is called photoelectric effect. b) When the zinc plate is exposed to UV light, electrons are emitted from the surface of the plate. This causes the plate to become negatively charged as electrons are leaving the surface. c) The rate of emission of electrons increases when the intensity of UV light is increased. This is because the intensity of the light determines the number of photons incident on the surface of the plate, and each photon can cause an electron to be emitted. d) The work function energy of a metal is the minimum energy required to remove an electron from the surface of the metal. In the case of zinc, it means that an energy of 4.24 eV or more is required to remove an electron from the surface of the zinc plate.

a) A photon is a quantum of electromagnetic radiation that carries energy and momentum. It behaves like a particle in certain interactions, but also exhibits wave-like properties. b) i. hf is the energy of a single photon, where h is Planck's constant and f is the frequency of the electromagnetic radiation. mvmax² is the maximum kinetic energy of an ejected electron, where m is the mass of the electron and vmax is its maximum speed. ii. The threshold frequency is the minimum frequency of radiation required to eject an electron from the surface of a metal. It can be calculated using the equation E = hf, where E is the work function energy of the metal. The threshold frequency is f = E/h = 1.9 eV / (6.626 x 10^-34 J s) = 2.86 x 10^15 Hz. iii. When the intensity of the incident radiation is doubled, the number of photons incident on the surface of the metal is doubled, which increases the number of ejected electrons. The maximum kinetic energy of each photoelectron does not change, as it depends only on the frequency of the radiation.

i.

The charge reaching the collector in 5.0 s is Q = It = (1.2 x 10^-7 A) x (5.0 s) = 6.0 x 10^-7 C.

The number of photoelectrons reaching the collector in 5.0 s can be calculated using the equation Q = ne, where n is the number of electrons and e is the elementary charge. Therefore, n = Q/e = (6.0 x 10^-7 C) / (1.602 x 10^-19 C/electron) = 3.74 x 10^12 electrons. ii. The maximum kinetic energy of a photoelectron can be calculated using Einstein's photoelectric equation: hf = φ + 1/2mv^2, where φ is the work function energy of the metal, m is the mass of the electron, v is its speed, and h is Planck's constant. Rearranging the equation to solve for v^2, we get v^2 = 2hf/m - 2φ/m. Plugging in the values, we get v^2 = (2 x 6.626 x 10^-34 J s x 7.0 x 10^14 Hz) / (9.109 x 10^-31 kg) - (2 x 3.5 x 10^-19 J) / (9.109 x 10^-31 kg) = 5.16 x 10^5 m^2/s^2. Therefore, the maximum kinetic energy of a photoelectron is KEmax = 1/2mv^2 = (1/2) x (9.109 x 10^-31 kg) x

c) For a particular wavelength of incident light, sodium releases photoelectrons. State how the rate of releases of photoelectrons changes with the intensity of light is doubled. Explain your answer.

When the intensity of the incident light is doubled, the rate of photoelectron emission from the sodium metal will also double. This is because the number of photons striking the surface of the metal and ejecting photoelectrons will increase with the intensity of the light. The rate of photoelectron emission is directly proportional to the number of photons absorbed by the metal, and therefore to the intensity of the incident light.

d) i. The terms hf and mvmax² in Einstein's photoelectric equation represent the energy of a single photon and the maximum kinetic energy of a photoelectron ejected from the metal, respectively. hf is the energy of the photon, where h is Planck's constant and f is the frequency of the incident radiation. mvmax² is the maximum kinetic energy of the photoelectron, where m is the mass of the electron and vmax is its maximum speed.

ii. To calculate the threshold frequency, we can use the formula:

hf = Φ + KE

where Φ is the work function energy and KE is the kinetic energy of the ejected photoelectron. At the threshold frequency, the kinetic energy is zero, so we have:

hf = Φ

Solving for f, we get:

f = Φ / h

Substituting the given values, we get:

f = (1.9 eV) / (4.14 x 10^-15 eV s) = 4.59 x 10^14 Hz

iii. When the intensity of the incident radiation is doubled, the number of photons striking the surface of the metal will double, but the energy of each photon will remain the same. As a result, the maximum kinetic energy of the ejected photoelectrons will also remain the same, but the rate of photoelectron emission will double, as explained in part c).

i. To calculate the charge reaching the collector in 5.0 s, we can use the formula:

Q = It

where Q is the charge, I is the current, and t is the time. Substituting the given values, we get: