(a) Photoelectrons with a maximum kinetic energy of 7.6eV are emitted from a metal when it is illuminated by ultraviolet radiation with a wavelength of 1.3×10^2 nm.
(i) What is the energy of the incident photons in electronvolts?
(ii) What is the wavelength of the radiation (in nanometres) corresponding to the lowest energy photons that can free electrons from the metal?
(iii) How do you explain the fact that, when infrared radiation is shone on this metal, no photoelectrons are emitted?
(b) What is the de Broglie wavelength (in nanometres) of an electron with a kinetic energy of 7.6eV?
(a) To answer these questions, we can use the following equations and principles:
(i) The energy of a photon can be calculated using the equation:
Energy of a photon = Planck's constant (h) x frequency (ν)
We can find the frequency using the equation:
Speed of light (c) = wavelength (λ) x frequency (ν)
Rearranging the equation, we get:
Frequency (ν) = Speed of light (c) / wavelength (λ)
Substituting the values into the equation, we have:
Frequency (ν) = (Speed of light) / (wavelength) = (3 x 10^8 m/s) / (1.3 x 10^-2 m)
= 2.308 x 10^10 Hz
Now, using the equation E = hν, we can find the energy of the incident photons:
Energy of incident photons = Planck's constant (h) x Frequency (ν)
= (6.63 x 10^-34 J·s) x (2.308 x 10^10 Hz)
= 1.53 x 10^-23 J
Converting this energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
Energy of incident photons = (1.53 x 10^-23 J) / (1.6 x 10^-19 J/eV)
= 9.56 eV
Therefore, the energy of the incident photons is 9.56 eV.
(ii) The energy of the lowest energy photons that can free electrons from the metal is equal to the work function (φ) of the metal. The work function is the minimum energy required for photoelectron emission.
Work Function (φ) = Maximum Kinetic energy of photoelectrons
In this case, the maximum kinetic energy of the photoelectrons is given as 7.6 eV.
Therefore, the wavelength (λ) of the radiation corresponding to the lowest energy photons is given by the equation:
Energy of a photon = Planck's constant (h) x frequency (ν)
Rearranging the equation, we have:
λ = (hc) / (Energy of a photon)
Substituting the values, we get:
λ = (6.63 x 10^-34 J·s x 3 x 10^8 m/s) / (7.6 x 1.6 x 10^-19 J)
λ ≈ 1.612 x 10^-7 m
Converting this wavelength to nanometers:
λ ≈ (1.612 x 10^-7 m) x (10^9 nm/m)
≈ 161.2 nm
Therefore, the wavelength of the radiation corresponding to the lowest energy photons is approximately 161.2 nm.
(iii) When infrared radiation is shone on the metal, no photoelectrons are emitted because the energy of the photons in the infrared region is lower than the work function (φ) of the metal. The work function is the minimum amount of energy required to remove electrons from the metal surface. Infrared photons do not possess enough energy to overcome the work function and free electrons from the metal.
(b) The de Broglie wavelength of an electron can be calculated using the equation:
λ = h / p,
where h is Planck's constant and p is the momentum of the electron. The momentum of an electron can be calculated using the formula:
p = √(2mK),
where m is the mass of the electron and K is the kinetic energy of the electron.
Given the kinetic energy as 7.6 eV, we need to convert it to joules:
1 eV = 1.6 x 10^-19 J
Kinetic energy (K) = 7.6 eV x (1.6 x 10^-19 J/eV)
= 1.216 x 10^-18 J
Now, we can calculate the momentum of the electron:
p = √(2mK)
= √(2 x (9.11 x 10^-31 kg) x (1.216 x 10^-18 J))
≈ 9.22 x 10^-23 kg·m/s
Finally, we can calculate the de Broglie wavelength using the equation:
λ = h / p
= (6.63 x 10^-34 J·s) / (9.22 x 10^-23 kg·m/s)
≈ 7.18 x 10^-12 m
Converting this wavelength to nanometers:
λ ≈ (7.18 x 10^-12 m) x (10^9 nm/m)
≈ 7.18 nm
Therefore, the de Broglie wavelength of an electron with a kinetic energy of 7.6 eV is approximately 7.18 nm.
To answer these questions, we need to apply the principles of photon energy and the photoelectric effect.
(a) Photoelectrons with a maximum kinetic energy of 7.6 eV are emitted from a metal when it is illuminated by ultraviolet radiation with a wavelength of 1.3 × 10^2 nm.
(i) To determine the energy of the incident photons in electronvolts, we can use the equation:
E = hc/λ
Where E is the energy of the photon, h is Planck's constant (6.626 × 10^-34 Js), c is the speed of light (3.0 × 10^8 m/s), and λ is the wavelength of radiation.
First, we need to convert the given wavelength from nanometers (nm) to meters (m):
λ = 1.3 × 10^2 nm = 1.3 × 10^-7 m
Now, we can calculate the energy of the incident photons:
E = (6.626 × 10^-34 Js)(3.0 × 10^8 m/s) / (1.3 × 10^-7 m)
E ≈ 4.82 × 10^-19 J
To convert this energy from joules to electronvolts (eV), we can use the conversion factor 1 eV = 1.6 × 10^-19 J:
E(eV) = (4.82 × 10^-19 J) / (1.6 × 10^-19 J/eV)
E(eV) ≈ 3.01 eV
Therefore, the energy of the incident photons is approximately 3.01 eV.
(ii) To determine the wavelength of the radiation corresponding to the lowest energy photons that can free electrons from the metal, we use the equation:
E = hc/λ
Since the photoelectrons have a maximum kinetic energy of 7.6 eV, the energy of the incident photons must exceed or equal this value.
Let's assume the energy of the incident photons is equal to the work function of the metal (ϕ) plus the maximum kinetic energy of the emitted electron (Ekmax).
Eincident = ϕ + Ekmax
Since Ekmax is given as 7.6 eV, we can subtract it from 3.01 eV to find the work function of the metal ϕ.
ϕ = Eincident - Ekmax
ϕ = 3.01 eV - 7.6 eV
ϕ ≈ -4.59 eV
A negative work function doesn't make physical sense. It indicates that no photons with a lower energy than the maximum kinetic energy of the emitted electrons will be able to free electrons from the metal.
Hence, there is no wavelength of radiation corresponding to the lowest energy photons that can free electrons from the metal.
(iii) When infrared radiation is shone on the metal, no photoelectrons are emitted because the energy of infrared photons is insufficient to overcome the work function of the metal. These photons do not have enough energy to excite the electrons in the metal to a level where they can be emitted.
(b) The de Broglie wavelength (λ) of an electron with a kinetic energy (KE) can be determined using the equation:
λ = h / √(2mKE)
Where h is Planck's constant and m is the mass of the electron.
First, we need to convert the kinetic energy from eV to joules:
KE(J) = 7.6 eV × (1.6 × 10^-19 J/eV)
KE(J) ≈ 1.22 × 10^-18 J
Now, we can calculate the de Broglie wavelength:
λ = (6.626 × 10^-34 Js) / √(2 × (9.109 × 10^-31 kg) × (1.22 × 10^-18 J))
λ ≈ 1.23 × 10^-10 m
To convert this wavelength from meters to nanometers, we multiply by 10^9:
λ(nm) ≈ 1.23 × 10^-10 m × 10^9 nm/m
λ(nm) ≈ 123 nm
Therefore, the de Broglie wavelength of an electron with a kinetic energy of 7.6 eV is approximately 123 nm.
https://physics.info/photoelectric/
See above link or Google Albert Einstein.
(i) get energy of incident photon from wavelength. E = h f but
f = c/wavelength
(ii)Ke emmitted = incident energy - energy needed to shake an electron loose
(iii) infrared is low frequency(energy)
http://calistry.org/calculate/deBroglieEquation