Answer :
Certainly. Let's tackle each part of the problem step-by-step.
### Part (a)
Explain what is meant by the term 'work function'.
The work function of a metal is the minimum amount of energy required to remove an electron from the surface of the metal. This energy must be provided by the incident radiation (photon energy). If the energy of the incident photon is equal to or greater than the work function of the metal, electrons can be emitted from the surface. This process is known as the photoelectric effect.
### Part (b)
Light with a frequency of [tex]\(1.2 \times 10^{15} \text{ Hz}\)[/tex] is shone onto the metal surface. Find whether or not the photons of this light will cause the photoelectric effect to take place. [tex]\(\left( h = 6.62 \times 10^{-34} \text{ Js} \right)\)[/tex]
1. Calculate the energy of one photon using the formula:
[tex]\[ E_{\text{photon}} = h \times \text{frequency} \][/tex]
2. Substitute the given values:
[tex]\[ E_{\text{photon}} = (6.62 \times 10^{-34} \text{ Js}) \times (1.2 \times 10^{15} \text{ Hz}) \][/tex]
3. Calculate:
[tex]\[ E_{\text{photon}} = 7.944 \times 10^{-19} \text{ J} \][/tex]
4. Compare the photon energy with the work function:
The energy of the photon is [tex]\(7.944 \times 10^{-19} \text{ J}\)[/tex] which is greater than the work function [tex]\(6.4 \times 10^{-19} \text{ J}\)[/tex]. Therefore, the photons of this light will cause the photoelectric effect to take place.
### Part (c)
The light source is now replaced with a light source of frequency [tex]\(1.5 \times 10^{15} \text{ Hz}\)[/tex].
#### (i)
Write down Einstein's photoelectric equation.
Einstein's photoelectric equation is:
[tex]\[ E_{\text{photon}} = \text{Work function} + \text{Kinetic Energy of ejected electrons} \][/tex]
where [tex]\( E_{\text{photon}} \)[/tex] is the energy of the incident photon.
#### (ii)
The photons from the source contain more energy than is required to release the electrons. How much extra energy is available after the electron has been released?
1. Calculate the energy of one photon for the new frequency:
[tex]\[ E_{\text{photon}} = h \times \text{frequency} \][/tex]
2. Substitute the given values:
[tex]\[ E_{\text{photon}} = (6.62 \times 10^{-34} \text{ Js}) \times (1.5 \times 10^{15} \text{ Hz}) \][/tex]
3. Calculate:
[tex]\[ E_{\text{photon}} = 9.93 \times 10^{-19} \text{ J} \][/tex]
4. Find the extra energy after the work function has been overcome:
[tex]\[ \text{Extra energy} = E_{\text{photon}} - \text{Work function} \][/tex]
[tex]\[ \text{Extra energy} = 9.93 \times 10^{-19} \text{ J} - 6.4 \times 10^{-19} \text{ J} \][/tex]
5. Calculate:
[tex]\[ \text{Extra energy} = 3.53 \times 10^{-19} \text{ J} \][/tex]
#### (iii)
Photons come from three lamps that emit red, green, and blue light. Which of these lamps produces photons with the highest energy?
The energy of a photon is directly proportional to its frequency. Among red, green, and blue light, blue light has the highest frequency. Therefore, photons from the blue lamp will have the highest energy.
### Part (a)
Explain what is meant by the term 'work function'.
The work function of a metal is the minimum amount of energy required to remove an electron from the surface of the metal. This energy must be provided by the incident radiation (photon energy). If the energy of the incident photon is equal to or greater than the work function of the metal, electrons can be emitted from the surface. This process is known as the photoelectric effect.
### Part (b)
Light with a frequency of [tex]\(1.2 \times 10^{15} \text{ Hz}\)[/tex] is shone onto the metal surface. Find whether or not the photons of this light will cause the photoelectric effect to take place. [tex]\(\left( h = 6.62 \times 10^{-34} \text{ Js} \right)\)[/tex]
1. Calculate the energy of one photon using the formula:
[tex]\[ E_{\text{photon}} = h \times \text{frequency} \][/tex]
2. Substitute the given values:
[tex]\[ E_{\text{photon}} = (6.62 \times 10^{-34} \text{ Js}) \times (1.2 \times 10^{15} \text{ Hz}) \][/tex]
3. Calculate:
[tex]\[ E_{\text{photon}} = 7.944 \times 10^{-19} \text{ J} \][/tex]
4. Compare the photon energy with the work function:
The energy of the photon is [tex]\(7.944 \times 10^{-19} \text{ J}\)[/tex] which is greater than the work function [tex]\(6.4 \times 10^{-19} \text{ J}\)[/tex]. Therefore, the photons of this light will cause the photoelectric effect to take place.
### Part (c)
The light source is now replaced with a light source of frequency [tex]\(1.5 \times 10^{15} \text{ Hz}\)[/tex].
#### (i)
Write down Einstein's photoelectric equation.
Einstein's photoelectric equation is:
[tex]\[ E_{\text{photon}} = \text{Work function} + \text{Kinetic Energy of ejected electrons} \][/tex]
where [tex]\( E_{\text{photon}} \)[/tex] is the energy of the incident photon.
#### (ii)
The photons from the source contain more energy than is required to release the electrons. How much extra energy is available after the electron has been released?
1. Calculate the energy of one photon for the new frequency:
[tex]\[ E_{\text{photon}} = h \times \text{frequency} \][/tex]
2. Substitute the given values:
[tex]\[ E_{\text{photon}} = (6.62 \times 10^{-34} \text{ Js}) \times (1.5 \times 10^{15} \text{ Hz}) \][/tex]
3. Calculate:
[tex]\[ E_{\text{photon}} = 9.93 \times 10^{-19} \text{ J} \][/tex]
4. Find the extra energy after the work function has been overcome:
[tex]\[ \text{Extra energy} = E_{\text{photon}} - \text{Work function} \][/tex]
[tex]\[ \text{Extra energy} = 9.93 \times 10^{-19} \text{ J} - 6.4 \times 10^{-19} \text{ J} \][/tex]
5. Calculate:
[tex]\[ \text{Extra energy} = 3.53 \times 10^{-19} \text{ J} \][/tex]
#### (iii)
Photons come from three lamps that emit red, green, and blue light. Which of these lamps produces photons with the highest energy?
The energy of a photon is directly proportional to its frequency. Among red, green, and blue light, blue light has the highest frequency. Therefore, photons from the blue lamp will have the highest energy.
