In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be

Options

A. `v(3/4)^(1/2)`
B. `v(4/3)^(1/2)`
C. less than `v(4/3)^(1/2)`
D. greater than `v (4/3)^(1/2)`

Answer

To solve the problem, we will analyze the photoelectric effect for two different wavelengths and derive the relationship between the speeds of the fastest electrons.

### Step-by-step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of the incident photons can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

2. **Energy Equation for the First Wavelength**:
For the initial wavelength \( \lambda \), the energy of the photon is given by:
\[ E_1 = \frac{hc}{\lambda} \]
According to the photoelectric effect, this energy is used to overcome the work function \( \phi \) of the metal and provide kinetic energy to the fastest electron:
\[ E_1 = \phi + \frac{1}{2} mv^2 \]
Rearranging gives:
\[ \frac{hc}{\lambda} = \phi + \frac{1}{2} mv^2 \quad \text{(Equation 1)} \]

3. **Energy Equation for the Second Wavelength**:
When the wavelength is changed to \( \frac{3\lambda}{4} \), the energy of the photon becomes:
\[ E_2 = \frac{hc}{\frac{3\lambda}{4}} = \frac{4hc}{3\lambda} \]
The energy equation now is:
\[ E_2 = \phi + \frac{1}{2} mv'^2 \]
Rearranging gives:
\[ \frac{4hc}{3\lambda} = \phi + \frac{1}{2} mv'^2 \quad \text{(Equation 2)} \]

4. **Subtracting the Two Equations**:
We will subtract Equation 1 from Equation 2:
\[ \frac{4hc}{3\lambda} - \frac{hc}{\lambda} = \left(\phi + \frac{1}{2} mv'^2\right) - \left(\phi + \frac{1}{2} mv^2\right) \]
Simplifying the left side:
\[ \frac{4hc}{3\lambda} - \frac{3hc}{3\lambda} = \frac{hc}{3\lambda} \]
The right side simplifies to:
\[ \frac{1}{2} mv'^2 - \frac{1}{2} mv^2 \]
Thus, we have:
\[ \frac{hc}{3\lambda} = \frac{1}{2} mv'^2 - \frac{1}{2} mv^2 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{hc}{3\lambda} = \frac{1}{2} m(v'^2 - v^2) \]
Multiplying through by 2:
\[ \frac{2hc}{3\lambda} = m(v'^2 - v^2) \]

6. **Assuming Work Function is Zero**:
For simplicity, if we assume \( \phi = 0 \):
\[ \frac{2hc}{3\lambda} = mv'^2 - mv^2 \]
This means:
\[ mv'^2 = mv^2 + \frac{2hc}{3\lambda} \]
Dividing by \( m \):
\[ v'^2 = v^2 + \frac{2hc}{3m\lambda} \]

7. **Finding the Relationship**:
Since \( \frac{hc}{\lambda} = \phi + \frac{1}{2} mv^2 \), we can see that the increase in energy due to the change in wavelength will lead to an increase in the speed of the fastest electron.

Thus, we can conclude that:
\[ v' > v \]

8. **Final Conclusion**:
The speed of the fastest electron when the wavelength is changed to \( \frac{3\lambda}{4} \) will be greater than \( v \).