In a photoelectric experiment, with light of wavelength `lamda` , the fastest electron has speed v. If the exciting wavelength is changed to `5lamda/4` , the speed of the fastest emitted electron will become

Options

A. `v sqrt(5/4)`
B. `v sqrt(5/3)`
C. less than `v sqrt(5/3)`
D. greater than `v sqrt(5/3)`

Explanation

To solve the problem step by step, we will use the principles of the photoelectric effect and the relationship between the kinetic energy of emitted electrons and the wavelength of the incident light.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The kinetic energy of the emitted electrons in a photoelectric experiment is given by the equation:
\[ K.E. = \frac{1}{2} mv^2 = E - \phi \]
where \(E\) is the energy of the incident photon, \(\phi\) is the work function of the material, and \(m\) is the mass of the electron.

2. **Calculate the Energy of the Photon**:
The energy of a photon can be expressed in terms of its wavelength \(\lambda\):
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light.

3. **Initial Conditions**:
For the initial wavelength \(\lambda\), the maximum kinetic energy of the emitted electrons is:
\[ K.E. = \frac{1}{2} mv^2 = \frac{hc}{\lambda} - \phi \]

4. **Change the Wavelength**:
When the wavelength is changed to \(\frac{5\lambda}{4}\), the new energy of the photon becomes:
\[ E' = \frac{hc}{\frac{5\lambda}{4}} = \frac{4hc}{5\lambda} \]

5. **New Kinetic Energy**:
The new maximum kinetic energy of the emitted electrons is:
\[ K.E.' = \frac{1}{2} mv'^2 = E' - \phi = \frac{4hc}{5\lambda} - \phi \]

6. **Set Up the Ratio of Kinetic Energies**:
We can now set up the ratio of the kinetic energies:
\[ \frac{K.E.'}{K.E.} = \frac{\frac{1}{2} mv'^2}{\frac{1}{2} mv^2} = \frac{v'^2}{v^2} \]

7. **Substituting the Energies**:
Substituting the expressions for kinetic energies:
\[ \frac{v'^2}{v^2} = \frac{\frac{4hc}{5\lambda} - \phi}{\frac{hc}{\lambda} - \phi} \]

8. **Simplifying the Expression**:
To simplify, we can express the right-hand side:
\[ \frac{v'^2}{v^2} = \frac{4\left(\frac{hc}{5\lambda} - \frac{\phi}{5}\right)}{hc - \phi} \]
This can be further simplified to:
\[ v'^2 = v^2 \cdot \frac{4}{5} \cdot \frac{hc - \phi}{hc - \phi + \frac{4\phi}{5}} \]

9. **Conclusion**:
Since the expression for \(v'^2\) is less than \(v^2\), we conclude that:
\[ v' < \sqrt{\frac{4}{5}} v \]

### Final Result:
The speed of the fastest emitted electron when the wavelength is changed to \(\frac{5\lambda}{4}\) will be less than \(\sqrt{\frac{4}{5}} v\).