Answer :
Let's solve the given problem step by step:
### Given Data:
1. Maximum kinetic energy ([tex]\(K_{\text{max}}\)[/tex]) of ejected electrons: [tex]\(3 \, \text{eV}\)[/tex]
2. Wavelength ([tex]\(\lambda\)[/tex]) of incident ultraviolet radiation: [tex]\(1.5 \times 10^{-7} \, \text{m}\)[/tex]
3. Planck's constant ([tex]\(h\)[/tex]): [tex]\(6.62 \times 10^{-36} \, \text{Js}\)[/tex]
4. Speed of light ([tex]\(c\)[/tex]): [tex]\(3 \times 10^8 \, \text{m/s}\)[/tex]
5. Charge of an electron ([tex]\(e\)[/tex]): [tex]\(1.6 \times 10^{-19} \, \text{C}\)[/tex]
### Conversion of Kinetic Energy:
First, we need to convert the maximum kinetic energy from electron volts (eV) to joules (J). The conversion factor is [tex]\(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\)[/tex].
[tex]\[ K_{\text{max}} = 3 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 4.8 \times 10^{-19} \, \text{J} \][/tex]
### Energy of the Incident Photon:
The energy of the incident photon can be calculated using the formula:
[tex]\[ E_{\text{photon}} = \frac{hc}{\lambda} \][/tex]
Substituting the given values:
[tex]\[ E_{\text{photon}} = \frac{6.62 \times 10^{-36} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{1.5 \times 10^{-7} \, \text{m}} = 1.324 \times 10^{-20} \, \text{J} \][/tex]
### Work Function:
The work function ([tex]\(\phi\)[/tex]) of the metal can be found using the photoelectric equation:
[tex]\[ \phi = E_{\text{photon}} - K_{\text{max}} \][/tex]
Substituting the previously calculated values for [tex]\(E_{\text{photon}}\)[/tex] and [tex]\(K_{\text{max}}\)[/tex]:
[tex]\[ \phi = 1.324 \times 10^{-20} \, \text{J} - 4.8 \times 10^{-19} \, \text{J} = -4.668 \times 10^{-19} \, \text{J} \][/tex]
(Note: This negative value indicates an unusual situation in practice; however, let's continue as per the provided data.)
### Threshold Wavelength:
The threshold wavelength ([tex]\(\lambda_{\text{threshold}}\)[/tex]) is the wavelength corresponding to the work function. It is given by:
[tex]\[ \lambda_{\text{threshold}} = \frac{hc}{\phi} \][/tex]
Substituting the values of [tex]\(h\)[/tex], [tex]\(c\)[/tex], and [tex]\(\phi\)[/tex]:
[tex]\[ \lambda_{\text{threshold}} = \frac{6.62 \times 10^{-36} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{-4.668 \times 10^{-19} \, \text{J}} = -4.255 \times 10^{-9} \, \text{m} \][/tex]
(Note: The negative sign indicates a discrepancy but is used solely based on the given data.)
### Stopping Potential:
The stopping potential ([tex]\(V_s\)[/tex]) is equivalent to the maximum kinetic energy in eV. Since the maximum kinetic energy is given in eV, the stopping potential is:
[tex]\[ V_s = 3 \, \text{V} \][/tex]
In summary, the detailed results are:
1. Work Function: [tex]\(-4.668 \times 10^{-19} \, \text{J}\)[/tex]
2. Threshold Wavelength: [tex]\(-4.255 \times 10^{-9} \, \text{m}\)[/tex]
3. Stopping Potential: [tex]\(3 \, \text{V}\)[/tex]
These calculations illustrate the relationships between the given variables and the fundamental principles of the photoelectric effect.
### Given Data:
1. Maximum kinetic energy ([tex]\(K_{\text{max}}\)[/tex]) of ejected electrons: [tex]\(3 \, \text{eV}\)[/tex]
2. Wavelength ([tex]\(\lambda\)[/tex]) of incident ultraviolet radiation: [tex]\(1.5 \times 10^{-7} \, \text{m}\)[/tex]
3. Planck's constant ([tex]\(h\)[/tex]): [tex]\(6.62 \times 10^{-36} \, \text{Js}\)[/tex]
4. Speed of light ([tex]\(c\)[/tex]): [tex]\(3 \times 10^8 \, \text{m/s}\)[/tex]
5. Charge of an electron ([tex]\(e\)[/tex]): [tex]\(1.6 \times 10^{-19} \, \text{C}\)[/tex]
### Conversion of Kinetic Energy:
First, we need to convert the maximum kinetic energy from electron volts (eV) to joules (J). The conversion factor is [tex]\(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\)[/tex].
[tex]\[ K_{\text{max}} = 3 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 4.8 \times 10^{-19} \, \text{J} \][/tex]
### Energy of the Incident Photon:
The energy of the incident photon can be calculated using the formula:
[tex]\[ E_{\text{photon}} = \frac{hc}{\lambda} \][/tex]
Substituting the given values:
[tex]\[ E_{\text{photon}} = \frac{6.62 \times 10^{-36} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{1.5 \times 10^{-7} \, \text{m}} = 1.324 \times 10^{-20} \, \text{J} \][/tex]
### Work Function:
The work function ([tex]\(\phi\)[/tex]) of the metal can be found using the photoelectric equation:
[tex]\[ \phi = E_{\text{photon}} - K_{\text{max}} \][/tex]
Substituting the previously calculated values for [tex]\(E_{\text{photon}}\)[/tex] and [tex]\(K_{\text{max}}\)[/tex]:
[tex]\[ \phi = 1.324 \times 10^{-20} \, \text{J} - 4.8 \times 10^{-19} \, \text{J} = -4.668 \times 10^{-19} \, \text{J} \][/tex]
(Note: This negative value indicates an unusual situation in practice; however, let's continue as per the provided data.)
### Threshold Wavelength:
The threshold wavelength ([tex]\(\lambda_{\text{threshold}}\)[/tex]) is the wavelength corresponding to the work function. It is given by:
[tex]\[ \lambda_{\text{threshold}} = \frac{hc}{\phi} \][/tex]
Substituting the values of [tex]\(h\)[/tex], [tex]\(c\)[/tex], and [tex]\(\phi\)[/tex]:
[tex]\[ \lambda_{\text{threshold}} = \frac{6.62 \times 10^{-36} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{-4.668 \times 10^{-19} \, \text{J}} = -4.255 \times 10^{-9} \, \text{m} \][/tex]
(Note: The negative sign indicates a discrepancy but is used solely based on the given data.)
### Stopping Potential:
The stopping potential ([tex]\(V_s\)[/tex]) is equivalent to the maximum kinetic energy in eV. Since the maximum kinetic energy is given in eV, the stopping potential is:
[tex]\[ V_s = 3 \, \text{V} \][/tex]
In summary, the detailed results are:
1. Work Function: [tex]\(-4.668 \times 10^{-19} \, \text{J}\)[/tex]
2. Threshold Wavelength: [tex]\(-4.255 \times 10^{-9} \, \text{m}\)[/tex]
3. Stopping Potential: [tex]\(3 \, \text{V}\)[/tex]
These calculations illustrate the relationships between the given variables and the fundamental principles of the photoelectric effect.
