Question

Photoelectric effect

7.
A metal surface having a work function of 3.0 eV is illuminated with radiation of wavelength
350nm. Calculate:
a) The threshold frequency (fo) and wavelength (Ao)
b) The maximum kinetic energy of the emitted photoelectrons
a) Calculate the work function (in eV) for a magnesium surface if the minimum frequency of
electromagnetic radiation which causes photoemission from the metal surface is
8.9 x 10¹4 Hz. in Joules
b) If the same surface were illuminated with radiation of wavelength 250 nm, calculate:
i. The maximum kinetic energy,
ii. The maximum velocity, of the emitted photoelectrons
9. When electromagnetic radiation of frequency 1.5 x 1014 Hz is incident on a metal surface, the
maximum kinetic energy of the emitted photoelectrons is found to be 3.8 x 10-20 J. Calculate the
work function of the metal.
10. Photons of electromagnetic radiation having energies of 1.0 eV, 2.0 eV and 4.0 eV are incident on a
metal surface having a work function of 1.7 eV.
a) Which of these photons will cause photoemission from the metal surface?
b) Calculate the maximum kinetic energies (in eV and J) of the liberated electrons in each of
those cases where photoemission occurs.
11. A vacuum photocell connected to a microammeter is illuminated with light of varying wavelength.
a) Explain why:
i. A photoelectric current is registered on the microammeter when light of a certain
wavelength is incident on the photocell.
ii. The current is found to increase with the light intensity is increased.
b) When the incident light wavelength is increased, the photoelectric current falls to zero. decre-
ased.
Explain why:
i. The current falls to zero.
ii. The current would still be zero if the light wavelength is kept the same and the
intensity is increased.

Answer 1

Step-by-step explanation:

7a) The work function (ϕ) is the minimum energy required to remove an electron from the metal surface. It is related to the threshold frequency (fo) by the equation:

ϕ = hfo

where h is Planck's constant (6.626 x 10^-34 J s).

The threshold wavelength (Ao) can be calculated from the threshold frequency using the equation:

c = λf

where c is the speed of light (3.00 x 10^8 m/s).

Given that the work function of the metal surface is 3.0 eV, we have:

ϕ = 3.0 eV = (3.0 x 1.6 x 10^-19) J fo = ϕ/h = (3.0 x 1.6 x 10^-19) J / (6.626 x 10^-34 J s) ≈ 4.53 x 10^14 Hz Ao = c/fo = (3.00 x 10^8 m/s) / (4.53 x 10^14 Hz) ≈ 661 nm

Therefore, the threshold frequency is 4.53 x 10^14 Hz and the threshold wavelength is approximately 661 nm.

7b) The maximum kinetic energy of the emitted photoelectrons can be calculated using the equation:

KEmax = hf - ϕ

where h is Planck's constant, f is the frequency of the incident radiation, and ϕ is the work function of the metal surface.

The energy of a photon can be calculated from its wavelength using the equation:

E = hc/λ

where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

Given that the wavelength of the incident radiation is 350 nm, we have:

f = c/λ = (3.00 x 10^8 m/s) / (350 x 10^-9 m) ≈ 8.57 x 10^14 Hz E = hc/λ = (6.626 x 10^-34 J s) x (3.00 x 10^8 m/s) / (350 x 10^-9 m) ≈ 1.79 eV

Therefore, the maximum kinetic energy of the emitted photoelectrons is:

KEmax = hf - ϕ = (6.626 x 10^-34 J s) x (8.57 x 10^14 Hz) - (3.0 x 1.6 x 10^-19) J ≈ 1.17 eV

a) The minimum frequency required to cause photoemission is equal to the threshold frequency:

fo = 8.9 x 10^14 Hz

Using the same equation as in part 7a), we can calculate the work function:

ϕ = hf0 = (6.626 x 10^-34 J s) x (8.9 x 10^14 Hz) ≈ 5.90 x 10^-19 J = 3.68 eV

b) i. The maximum kinetic energy of the emitted photoelectrons can be calculated using the same equation as in part 7b):

KEmax = hf - ϕ

The energy of a photon with wavelength 250 nm is:

E = hc/λ = (6.626 x 10^-34 J s) x (3.00 x 10^8 m/s) / (250 x 10^-9 m) ≈ 4.97 eV

Therefore, the maximum kinetic energy of the emitted photoelectrons is:

KEmax = hf -