The maximum kinetic energy of the photoelectrons is found to be `6.63xx10^(-19)J`, when the metal is irradiated with a radiation of frequency `2xx10^(15)`Hz. The threshold frequency of the metal is about:

Options

A. `1xx10^(15)s`
B. `1xx10^(15)s^(-1)`
C. `2.5xx10^(15)s^(-1)`
D. `4xx10^(15)s^(-1)`

Explanation

To find the threshold frequency of the metal, we will use the photoelectric effect equation derived from Einstein's theory. The equation relates the maximum kinetic energy (KE) of the emitted photoelectrons to the energy of the incident photons and the work function (W) of the metal.

### Step-by-step Solution:

1. **Write down the equation for the photoelectric effect**:
\[ KE = E - W \]
where \(KE\) is the kinetic energy of the emitted electrons, \(E\) is the energy of the incident photons, and \(W\) is the work function of the metal.

2. **Express the energy of the incident photons**:
The energy of the incident photons can be expressed using Planck's equation:
\[ E = h \nu \]
where \(h\) is Planck's constant and \(\nu\) is the frequency of the incident radiation.

3. **Express the work function**:
The work function can also be expressed in terms of the threshold frequency (\(\nu_0\)):
\[ W = h \nu_0 \]

4. **Substitute the expressions for \(E\) and \(W\) into the photoelectric equation**:
\[ KE = h \nu - h \nu_0 \]
Rearranging this gives:
\[ KE = h (\nu - \nu_0) \]

5. **Plug in the known values**:
We know:
- Maximum kinetic energy \(KE = 6.63 \times 10^{-19} \, J\)
- Frequency \(\nu = 2 \times 10^{15} \, Hz\)
- Planck's constant \(h = 6.63 \times 10^{-34} \, J \cdot s\)

Substituting these values into the equation:
\[ 6.63 \times 10^{-19} = 6.63 \times 10^{-34} (2 \times 10^{15} - \nu_0) \]

6. **Divide both sides by Planck's constant \(h\)**:
\[ \frac{6.63 \times 10^{-19}}{6.63 \times 10^{-34}} = 2 \times 10^{15} - \nu_0 \]
This simplifies to:
\[ 10^{15} = 2 \times 10^{15} - \nu_0 \]

7. **Rearranging to find the threshold frequency \(\nu_0\)**:
\[ \nu_0 = 2 \times 10^{15} - 10^{15} \]
\[ \nu_0 = 10^{15} \, Hz \]

### Final Answer:
The threshold frequency of the metal is approximately:
\[ \nu_0 = 1 \times 10^{15} \, Hz \]