Radiation of wavelength `lambda` in indent on a photocell . The fastest emitted electron has speed `v` if the wavelength is changed to `(3 lambda)/(4)` , then speed of the fastest emitted electron will be

Options

A. `lt v ((4)/(3))^((1)/(2))`
B. `= v ((4)/(3))^((1)/(2))`
C. `= v ((3)/(4))^((1)/(2))`
D. `gt v ((4)/(3))^((1)/(2))`

Answer

To solve the problem, we will use the photoelectric effect principles, specifically Einstein's photoelectric equation. The equation relates the energy of the incident photons to the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- Let the initial wavelength be \( \lambda \).
- The energy of the incident radiation can be expressed as:
\[ E_1 = \frac{hc}{\lambda} \]
- The kinetic energy of the fastest emitted electron is given by:
\[ K.E. = \frac{1}{2}mv^2 \]
- According to Einstein's photoelectric equation:
\[ \frac{hc}{\lambda} = \frac{1}{2}mv^2 + \phi \]
where \( \phi \) is the work function.

2. **Setting Up the Equation for the First Case:**
- Rearranging the equation gives:
\[ \frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi \tag{1} \]

3. **Changing the Wavelength:**
- Now, the wavelength is changed to \( \frac{3\lambda}{4} \).
- The energy of the incident radiation in this case is:
\[ E_2 = \frac{hc}{\frac{3\lambda}{4}} = \frac{4hc}{3\lambda} \]
- The new kinetic energy of the emitted electron can be expressed as:
\[ K.E. = \frac{1}{2}mv_2^2 \]
- Again, applying Einstein's photoelectric equation:
\[ \frac{4hc}{3\lambda} = \frac{1}{2}mv_2^2 + \phi \]

4. **Setting Up the Equation for the Second Case:**
- Rearranging this equation gives:
\[ \frac{1}{2}mv_2^2 = \frac{4hc}{3\lambda} - \phi \tag{2} \]

5. **Equating Work Functions from Both Cases:**
- From equations (1) and (2), we can equate the expressions for \( \phi \):
\[ \frac{1}{2}mv^2 + \phi = \frac{4hc}{3\lambda} - \phi \]
- Therefore, we can express \( \phi \) from equation (1):
\[ \phi = \frac{hc}{\lambda} - \frac{1}{2}mv^2 \]
- Substituting this into equation (2):
\[ \frac{1}{2}mv_2^2 = \frac{4hc}{3\lambda} - \left(\frac{hc}{\lambda} - \frac{1}{2}mv^2\right) \]
- Simplifying gives:
\[ \frac{1}{2}mv_2^2 = \frac{4hc}{3\lambda} - \frac{hc}{\lambda} + \frac{1}{2}mv^2 \]
- Combining terms:
\[ \frac{1}{2}mv_2^2 = \left(\frac{4hc}{3\lambda} - \frac{3hc}{3\lambda}\right) + \frac{1}{2}mv^2 \]
\[ \frac{1}{2}mv_2^2 = \frac{hc}{3\lambda} + \frac{1}{2}mv^2 \]

6. **Finding the Ratio of Speeds:**
- Rearranging gives:
\[ \frac{1}{2}mv_2^2 - \frac{1}{2}mv^2 = \frac{hc}{3\lambda} \]
- This implies that:
\[ v_2^2 = v^2 + \frac{2hc}{3m\lambda} \]
- Taking the square root gives:
\[ v_2 = \sqrt{v^2 + \frac{2hc}{3m\lambda}} \]

7. **Conclusion:**
- The new speed \( v_2 \) of the fastest emitted electron when the wavelength is changed to \( \frac{3\lambda}{4} \) is greater than \( v \) and can be expressed in terms of the original speed and the constants involved.

### Final Answer:
The speed of the fastest emitted electron when the wavelength is changed to \( \frac{3\lambda}{4} \) is:
\[ v_2 = \sqrt{v^2 + \frac{2hc}{3m\lambda}} \]