For photoelectric emission from certain metal the cut - off frequency is `v`. If radiation of frequency `2 v` incident on the metal plate , the maximum possible velocity of the emitted electron will be (`m` is the electron mass).

Options

A. (a)`(sqrt(hv)/(2m))`
B. (b)`(sqrt((2hv)/m))`
C. (c) `sqrt(2hv)/m`
D. (d) `2(sqrt(hv)/m)`

Answer

To solve the problem of finding the maximum possible velocity of emitted electrons when radiation of frequency \(2v\) is incident on a metal with a cut-off frequency \(v\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Work Function**:
The work function \(\phi\) of the metal is defined as the minimum energy required to remove an electron from the metal surface. It can be expressed in terms of the cut-off frequency \(v\):
\[ \phi = h v \]
where \(h\) is Planck's constant.

2. **Energy of Incident Radiation**:
The energy \(E\) of the incident radiation with frequency \(2v\) can be calculated using the formula:
\[ E = h \cdot (2v) = 2h v \]

3. **Apply Einstein's Photoelectric Equation**:
According to Einstein's photoelectric equation, the kinetic energy \(K.E.\) of the emitted electrons can be expressed as:
\[ K.E. = E - \phi \]
Substituting the expressions for \(E\) and \(\phi\):
\[ K.E. = 2h v - h v = h v \]

4. **Relate Kinetic Energy to Velocity**:
The kinetic energy of an emitted electron is also given by:
\[ K.E. = \frac{1}{2} m v_{max}^2 \]
where \(m\) is the mass of the electron and \(v_{max}\) is the maximum velocity of the emitted electron. Setting the two expressions for kinetic energy equal gives us:
\[ \frac{1}{2} m v_{max}^2 = h v \]

5. **Solve for Maximum Velocity**:
Rearranging the equation to solve for \(v_{max}\):
\[ v_{max}^2 = \frac{2h v}{m} \]
Taking the square root of both sides:
\[ v_{max} = \sqrt{\frac{2h v}{m}} \]

### Final Answer:
The maximum possible velocity of the emitted electron is:
\[ v_{max} = \sqrt{\frac{2h v}{m}} \]