Answer :
Sure, let's solve the problem step by step.
We can use the photoelectric equation which relates the kinetic energy (KE) of the emitted electron, the energy of the incident photons and the work function ([tex]\(\phi\)[/tex]) of the material. The equation is given by:
[tex]\[ \text{KE} = hf - \phi \][/tex]
Where:
- [tex]\( \text{KE} \)[/tex] is the kinetic energy of the emitted electron,
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( h \approx 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \)[/tex]),
- [tex]\( f \)[/tex] is the frequency of the incident light,
- [tex]\( \phi \)[/tex] is the work function of the material.
The frequency [tex]\( f \)[/tex] can be related to the wavelength [tex]\( \lambda \)[/tex] of the incident light by the equation:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
Where [tex]\( c \)[/tex] is the speed of light ([tex]\( c \approx 3 \times 10^8 \, \text{m/s} \)[/tex]).
For the first case:
- Let [tex]\( \lambda_1 \)[/tex] be the wavelength of the incident light,
- The kinetic energy of the emitted electrons is [tex]\( \text{KE}_1 = 1.8 \, \text{eV} \)[/tex].
Using the photoelectric equation, we can write:
[tex]\[ \text{KE}_1 = hf_1 - \phi \][/tex]
Substitute [tex]\( f_1 = \frac{c}{\lambda_1} \)[/tex]:
[tex]\[ 1.8 = \frac{hc}{\lambda_1} - \phi \quad (1) \][/tex]
For the second case:
- The wavelength of the incident light is reduced to [tex]\( \frac{\lambda_1}{2} \)[/tex],
- The kinetic energy of the emitted electrons is [tex]\( \text{KE}_2 = 5.5 \, \text{eV} \)[/tex].
Using the photoelectric equation again:
[tex]\[ \text{KE}_2 = hf_2 - \phi \][/tex]
Substitute [tex]\( f_2 = \frac{c}{\lambda_2} = \frac{c}{\frac{\lambda_1}{2}} = \frac{2c}{\lambda_1} \)[/tex]:
[tex]\[ 5.5 = \frac{2hc}{\lambda_1} - \phi \quad (2) \][/tex]
Now we have two equations:
1. [tex]\( 1.8 = \frac{hc}{\lambda_1} - \phi \)[/tex]
2. [tex]\( 5.5 = \frac{2hc}{\lambda_1} - \phi \)[/tex]
We can solve these equations simultaneously to find the work function [tex]\( \phi \)[/tex] and [tex]\( \lambda_1 \)[/tex].
First, subtract equation (1) from equation (2):
[tex]\[ 5.5 - 1.8 = \frac{2hc}{\lambda_1} - \phi - \left( \frac{hc}{\lambda_1} - \phi \right) \][/tex]
Simplify:
[tex]\[ 3.7 = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_1} \][/tex]
[tex]\[ 3.7 = \frac{hc}{\lambda_1} \][/tex]
Now solve for [tex]\( \lambda_1 \)[/tex]:
[tex]\[ \lambda_1 = \frac{hc}{3.7} \][/tex]
Substitute the values of [tex]\( h \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ \lambda_1 = \frac{4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{3.7 \, \text{eV}} \][/tex]
[tex]\[ \lambda_1 = \frac{1.2407 \times 10^{-6} \, \text{eV} \cdot \text{m}}{3.7 \, \text{eV}} \][/tex]
[tex]\[ \lambda_1 \approx 3.35 \times 10^{-7} \, \text{m} \][/tex]
[tex]\[ \lambda_1 \approx 335 \, \text{nm} \][/tex]
Now, substitute [tex]\( \lambda_1 \)[/tex] back into equation (1) to find [tex]\( \phi \)[/tex]:
[tex]\[ 1.8 = \frac{hc}{\lambda_1} - \phi \][/tex]
[tex]\[ 1.8 = \frac{4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{3.35 \times 10^{-7} \, \text{m}} - \phi \][/tex]
[tex]\[ 1.8 = \frac{1.2407 \times 10^{-6} \, \text{eV} \cdot \text{m}}{3.35 \times 10^{-7} \, \text{m}} - \phi \][/tex]
[tex]\[ 1.8 = 3.7 - \phi \][/tex]
[tex]\[ \phi = 3.7 - 1.8 \][/tex]
[tex]\[ \phi = 1.9 \, \text{eV} \][/tex]
Therefore, the work function of the cathode material is [tex]\( 1.9 \, \text{eV} \)[/tex].
We can use the photoelectric equation which relates the kinetic energy (KE) of the emitted electron, the energy of the incident photons and the work function ([tex]\(\phi\)[/tex]) of the material. The equation is given by:
[tex]\[ \text{KE} = hf - \phi \][/tex]
Where:
- [tex]\( \text{KE} \)[/tex] is the kinetic energy of the emitted electron,
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( h \approx 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \)[/tex]),
- [tex]\( f \)[/tex] is the frequency of the incident light,
- [tex]\( \phi \)[/tex] is the work function of the material.
The frequency [tex]\( f \)[/tex] can be related to the wavelength [tex]\( \lambda \)[/tex] of the incident light by the equation:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
Where [tex]\( c \)[/tex] is the speed of light ([tex]\( c \approx 3 \times 10^8 \, \text{m/s} \)[/tex]).
For the first case:
- Let [tex]\( \lambda_1 \)[/tex] be the wavelength of the incident light,
- The kinetic energy of the emitted electrons is [tex]\( \text{KE}_1 = 1.8 \, \text{eV} \)[/tex].
Using the photoelectric equation, we can write:
[tex]\[ \text{KE}_1 = hf_1 - \phi \][/tex]
Substitute [tex]\( f_1 = \frac{c}{\lambda_1} \)[/tex]:
[tex]\[ 1.8 = \frac{hc}{\lambda_1} - \phi \quad (1) \][/tex]
For the second case:
- The wavelength of the incident light is reduced to [tex]\( \frac{\lambda_1}{2} \)[/tex],
- The kinetic energy of the emitted electrons is [tex]\( \text{KE}_2 = 5.5 \, \text{eV} \)[/tex].
Using the photoelectric equation again:
[tex]\[ \text{KE}_2 = hf_2 - \phi \][/tex]
Substitute [tex]\( f_2 = \frac{c}{\lambda_2} = \frac{c}{\frac{\lambda_1}{2}} = \frac{2c}{\lambda_1} \)[/tex]:
[tex]\[ 5.5 = \frac{2hc}{\lambda_1} - \phi \quad (2) \][/tex]
Now we have two equations:
1. [tex]\( 1.8 = \frac{hc}{\lambda_1} - \phi \)[/tex]
2. [tex]\( 5.5 = \frac{2hc}{\lambda_1} - \phi \)[/tex]
We can solve these equations simultaneously to find the work function [tex]\( \phi \)[/tex] and [tex]\( \lambda_1 \)[/tex].
First, subtract equation (1) from equation (2):
[tex]\[ 5.5 - 1.8 = \frac{2hc}{\lambda_1} - \phi - \left( \frac{hc}{\lambda_1} - \phi \right) \][/tex]
Simplify:
[tex]\[ 3.7 = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_1} \][/tex]
[tex]\[ 3.7 = \frac{hc}{\lambda_1} \][/tex]
Now solve for [tex]\( \lambda_1 \)[/tex]:
[tex]\[ \lambda_1 = \frac{hc}{3.7} \][/tex]
Substitute the values of [tex]\( h \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ \lambda_1 = \frac{4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{3.7 \, \text{eV}} \][/tex]
[tex]\[ \lambda_1 = \frac{1.2407 \times 10^{-6} \, \text{eV} \cdot \text{m}}{3.7 \, \text{eV}} \][/tex]
[tex]\[ \lambda_1 \approx 3.35 \times 10^{-7} \, \text{m} \][/tex]
[tex]\[ \lambda_1 \approx 335 \, \text{nm} \][/tex]
Now, substitute [tex]\( \lambda_1 \)[/tex] back into equation (1) to find [tex]\( \phi \)[/tex]:
[tex]\[ 1.8 = \frac{hc}{\lambda_1} - \phi \][/tex]
[tex]\[ 1.8 = \frac{4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{3.35 \times 10^{-7} \, \text{m}} - \phi \][/tex]
[tex]\[ 1.8 = \frac{1.2407 \times 10^{-6} \, \text{eV} \cdot \text{m}}{3.35 \times 10^{-7} \, \text{m}} - \phi \][/tex]
[tex]\[ 1.8 = 3.7 - \phi \][/tex]
[tex]\[ \phi = 3.7 - 1.8 \][/tex]
[tex]\[ \phi = 1.9 \, \text{eV} \][/tex]
Therefore, the work function of the cathode material is [tex]\( 1.9 \, \text{eV} \)[/tex].
