Answer :
### Step-by-Step Solution:
### 10.1 Definition of Work Function
The work function of a metal is the minimum amount of energy required to eject an electron from the surface of the metal. It represents the energy needed to overcome the attractive forces binding the electron to the metal.
### 10.2 Calculation of the Work Function
We will use Einstein's photoelectric equation which is given by:
[tex]\[ KE_{max} = hf - \phi \][/tex]
Where:
- [tex]\( KE_{max} \)[/tex] is the maximum kinetic energy of the emitted electrons.
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( h = 6.626 \times 10^{-34} \)[/tex] Js).
- [tex]\( f \)[/tex] is the frequency of the incident light.
- [tex]\( \phi \)[/tex] is the work function of the metal.
First convert the given maximum kinetic energies and calculate the incident frequencies for both cases. Since [tex]\( f = \frac{c}{\lambda} \)[/tex]:
[tex]\[ f = \frac{c}{\lambda} \text{ where } c = 3 \times 10^8 \text{ m/s} \][/tex]
1. Initial Conditions:
[tex]\[ KE_{\text{initial}} = 4.48 \times 10^{-19} \text{ J} \][/tex]
2. Final Conditions when wavelength is increased by 50% ( [tex]\( \lambda \rightarrow 1.5 \lambda \)[/tex] ):
[tex]\[ KE_{\text{final}} = 1.76 \times 10^{-19} \text{ J} \][/tex]
Since wavelength increases by 50%, the frequency decreases by 33% as [tex]\( f = \frac{c}{\lambda} \)[/tex], therefore:
[tex]\[ f_{\text{final}} = \frac{f_{\text{initial}}}{1.5} \][/tex]
Using Einstein's photoelectric equation:
1. For Initial Conditions:
[tex]\[ 4.48 \times 10^{-19} \text{ J} = hf_{\text{initial}} - \phi \][/tex]
2. For Final Conditions:
[tex]\[ 1.76 \times 10^{-19} \text{ J} = h \left( \frac{f_{\text{initial}}}{1.5} \right) - \phi \][/tex]
From the second equation:
[tex]\[ 1.76 \times 10^{-19} \text{ J} = \frac{hf_{\text{initial}}}{1.5} - \phi \][/tex]
Now, let's solve the equations:
1. Multiply the second equation by 1.5:
[tex]\[ 1.5 \times 1.76 \times 10^{-19} = hf_{\text{initial}} - 1.5\phi \][/tex]
[tex]\[ 2.64 \times 10^{-19} = hf_{\text{initial}} - 1.5\phi \][/tex]
2. Subtract the first equation from this result:
[tex]\[ (2.64 \times 10^{-19}) - (4.48 \times 10^{-19}) = (hf_{\text{initial}} - 1.5\phi) - (hf_{\text{initial}} - \phi) \][/tex]
[tex]\[ -1.84 \times 10^{-19} = -0.5\phi \][/tex]
Solve for [tex]\(\phi\)[/tex]:
[tex]\[ \phi = 3.68 \times 10^{-19} \text{ J} \][/tex]
Thus, the work function [tex]\(\phi\)[/tex] is:
[tex]\[ \boxed{3.68 \times 10^{-19} \text{ J}} \][/tex]
### 10.3 Calculation of the Initial Wavelength
Using the calculated work function and the initial maximum kinetic energy:
[tex]\[ KE_{\text{initial}} = hf_{\text{initial}} - \phi \][/tex]
[tex]\[ 4.48 \times 10^{-19} \text{ J} = hf_{\text{initial}} - 3.68 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ hf_{\text{initial}} = 4.48 \times 10^{-19} \text{ J} + 3.68 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ hf_{\text{initial}} = 8.16 \times 10^{-19} \text{ J} \][/tex]
Converting [tex]\( hf_{\text{initial}} \)[/tex] to wavelength:
[tex]\[ \lambda_{\text{initial}} = \frac{hc}{hf_{\text{initial}}} \][/tex]
[tex]\[ \lambda_{\text{initial}} = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{8.16 \times 10^{-19} \text{ J}} \][/tex]
[tex]\[ \lambda_{\text{initial}} = 2.43 \times 10^{-7} \text{ m} \][/tex]
Thus, the initial wavelength is:
[tex]\[ \boxed{2.43 \times 10^{-7} \text{ m}} \text{ or } 243 \text{ nm} \][/tex]
### 10.4 Effect on Maximum Kinetic Energy
10.4.1 Using incident light of shorter wavelength:
- INCREASES (Shorter wavelength means higher frequency, thus more energy to the electrons)
10.4.2 Using a photocathode with a higher work function:
- DECREASES (More energy needed to remove electrons, so less kinetic energy left)
10.4.3 Using incident light of greater intensity:
- REMAINS THE SAME (Intensity affects the number of electrons ejected, not their kinetic energy)
### Summary
10.1 Definition of the term work function \- 2 points
10.2 Proof that the work function is [tex]\( 3.68 \times 10^{-19} \text{ J}\)[/tex] \- 6 points
10.3 Initial wavelength of the incident light \- 2 points
10.4 Effects on maximum kinetic energy \- 3 points
[tex]\[ \boxed{13 \text{ points}} \][/tex]
### 10.1 Definition of Work Function
The work function of a metal is the minimum amount of energy required to eject an electron from the surface of the metal. It represents the energy needed to overcome the attractive forces binding the electron to the metal.
### 10.2 Calculation of the Work Function
We will use Einstein's photoelectric equation which is given by:
[tex]\[ KE_{max} = hf - \phi \][/tex]
Where:
- [tex]\( KE_{max} \)[/tex] is the maximum kinetic energy of the emitted electrons.
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( h = 6.626 \times 10^{-34} \)[/tex] Js).
- [tex]\( f \)[/tex] is the frequency of the incident light.
- [tex]\( \phi \)[/tex] is the work function of the metal.
First convert the given maximum kinetic energies and calculate the incident frequencies for both cases. Since [tex]\( f = \frac{c}{\lambda} \)[/tex]:
[tex]\[ f = \frac{c}{\lambda} \text{ where } c = 3 \times 10^8 \text{ m/s} \][/tex]
1. Initial Conditions:
[tex]\[ KE_{\text{initial}} = 4.48 \times 10^{-19} \text{ J} \][/tex]
2. Final Conditions when wavelength is increased by 50% ( [tex]\( \lambda \rightarrow 1.5 \lambda \)[/tex] ):
[tex]\[ KE_{\text{final}} = 1.76 \times 10^{-19} \text{ J} \][/tex]
Since wavelength increases by 50%, the frequency decreases by 33% as [tex]\( f = \frac{c}{\lambda} \)[/tex], therefore:
[tex]\[ f_{\text{final}} = \frac{f_{\text{initial}}}{1.5} \][/tex]
Using Einstein's photoelectric equation:
1. For Initial Conditions:
[tex]\[ 4.48 \times 10^{-19} \text{ J} = hf_{\text{initial}} - \phi \][/tex]
2. For Final Conditions:
[tex]\[ 1.76 \times 10^{-19} \text{ J} = h \left( \frac{f_{\text{initial}}}{1.5} \right) - \phi \][/tex]
From the second equation:
[tex]\[ 1.76 \times 10^{-19} \text{ J} = \frac{hf_{\text{initial}}}{1.5} - \phi \][/tex]
Now, let's solve the equations:
1. Multiply the second equation by 1.5:
[tex]\[ 1.5 \times 1.76 \times 10^{-19} = hf_{\text{initial}} - 1.5\phi \][/tex]
[tex]\[ 2.64 \times 10^{-19} = hf_{\text{initial}} - 1.5\phi \][/tex]
2. Subtract the first equation from this result:
[tex]\[ (2.64 \times 10^{-19}) - (4.48 \times 10^{-19}) = (hf_{\text{initial}} - 1.5\phi) - (hf_{\text{initial}} - \phi) \][/tex]
[tex]\[ -1.84 \times 10^{-19} = -0.5\phi \][/tex]
Solve for [tex]\(\phi\)[/tex]:
[tex]\[ \phi = 3.68 \times 10^{-19} \text{ J} \][/tex]
Thus, the work function [tex]\(\phi\)[/tex] is:
[tex]\[ \boxed{3.68 \times 10^{-19} \text{ J}} \][/tex]
### 10.3 Calculation of the Initial Wavelength
Using the calculated work function and the initial maximum kinetic energy:
[tex]\[ KE_{\text{initial}} = hf_{\text{initial}} - \phi \][/tex]
[tex]\[ 4.48 \times 10^{-19} \text{ J} = hf_{\text{initial}} - 3.68 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ hf_{\text{initial}} = 4.48 \times 10^{-19} \text{ J} + 3.68 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ hf_{\text{initial}} = 8.16 \times 10^{-19} \text{ J} \][/tex]
Converting [tex]\( hf_{\text{initial}} \)[/tex] to wavelength:
[tex]\[ \lambda_{\text{initial}} = \frac{hc}{hf_{\text{initial}}} \][/tex]
[tex]\[ \lambda_{\text{initial}} = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{8.16 \times 10^{-19} \text{ J}} \][/tex]
[tex]\[ \lambda_{\text{initial}} = 2.43 \times 10^{-7} \text{ m} \][/tex]
Thus, the initial wavelength is:
[tex]\[ \boxed{2.43 \times 10^{-7} \text{ m}} \text{ or } 243 \text{ nm} \][/tex]
### 10.4 Effect on Maximum Kinetic Energy
10.4.1 Using incident light of shorter wavelength:
- INCREASES (Shorter wavelength means higher frequency, thus more energy to the electrons)
10.4.2 Using a photocathode with a higher work function:
- DECREASES (More energy needed to remove electrons, so less kinetic energy left)
10.4.3 Using incident light of greater intensity:
- REMAINS THE SAME (Intensity affects the number of electrons ejected, not their kinetic energy)
### Summary
10.1 Definition of the term work function \- 2 points
10.2 Proof that the work function is [tex]\( 3.68 \times 10^{-19} \text{ J}\)[/tex] \- 6 points
10.3 Initial wavelength of the incident light \- 2 points
10.4 Effects on maximum kinetic energy \- 3 points
[tex]\[ \boxed{13 \text{ points}} \][/tex]
