In a photomissive cell with executing wavelength `lamda//2` , work function `phi` , the fastest moving electron has acquired speed as v. If the executing wavelength is changed to `(5lamda)/4` , the speed of the fastest emitted electron will be

Options

A. `sqrt((8v^2)/5-(6phi)/(5m))`
B. `sqrt(v^2-(12hc)/(5lamdam))`
C. `sqrt((18v^2)/5-(6phi)/(5m))`
D. `sqrt((18mv^2)/5-(6phi)/(5m))`

Explanation

To solve the problem, we will use the photoelectric effect equations. The key is to relate the kinetic energy of the emitted electrons to the energy of the incident photons.

### Step-by-step Solution:

1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed using the equation:
\[ K.E. = E - \phi \]
where \(E\) is the energy of the incident photon and \(\phi\) is the work function.

2. **Photon Energy Calculation**:
The energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the incident light.

3. **Case 1: Initial Wavelength**:
For the initial wavelength \(\frac{\lambda}{2}\):
\[ E_1 = \frac{hc}{\frac{\lambda}{2}} = \frac{2hc}{\lambda} \]
The maximum kinetic energy of the emitted electrons is:
\[ K.E. = \frac{2hc}{\lambda} - \phi \]
This can also be expressed in terms of the speed \(v\) of the fastest emitted electron:
\[ K.E. = \frac{1}{2} mv^2 \]
Therefore, we have:
\[ \frac{2hc}{\lambda} - \phi = \frac{1}{2} mv^2 \quad \text{(Equation 1)} \]

4. **Case 2: New Wavelength**:
For the new wavelength \(\frac{5\lambda}{4}\):
\[ E_2 = \frac{hc}{\frac{5\lambda}{4}} = \frac{4hc}{5\lambda} \]
The maximum kinetic energy in this case is:
\[ K.E. = \frac{4hc}{5\lambda} - \phi \]
This can also be expressed in terms of the new speed \(v_1\):
\[ K.E. = \frac{1}{2} mv_1^2 \quad \text{(Equation 2)} \]

5. **Setting Up the Equations**:
From Equation 1 and Equation 2, we can set them equal:
\[ \frac{2hc}{\lambda} - \phi = \frac{1}{2} mv^2 \]
\[ \frac{4hc}{5\lambda} - \phi = \frac{1}{2} mv_1^2 \]

6. **Subtracting the Equations**:
Subtract the second equation from the first:
\[ \left(\frac{2hc}{\lambda} - \phi\right) - \left(\frac{4hc}{5\lambda} - \phi\right) = \frac{1}{2} mv^2 - \frac{1}{2} mv_1^2 \]
Simplifying:
\[ \frac{2hc}{\lambda} - \frac{4hc}{5\lambda} = \frac{1}{2} m(v^2 - v_1^2) \]
Finding a common denominator:
\[ \frac{10hc - 4hc}{5\lambda} = \frac{1}{2} m(v^2 - v_1^2) \]
\[ \frac{6hc}{5\lambda} = \frac{1}{2} m(v^2 - v_1^2) \]

7. **Rearranging for \(v_1\)**:
Rearranging gives:
\[ v^2 - v_1^2 = \frac{12hc}{5m\lambda} \]
Thus:
\[ v_1^2 = v^2 - \frac{12hc}{5m\lambda} \]
Taking the square root:
\[ v_1 = \sqrt{v^2 - \frac{12hc}{5m\lambda}} \]

### Final Answer:
The speed of the fastest emitted electron when the wavelength is changed to \(\frac{5\lambda}{4}\) is:
\[ v_1 = \sqrt{v^2 - \frac{12hc}{5m\lambda}} \]