A certain metal when irradiated to light `(v = 3.2 xx 10^(16) Hz)` emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiation by light `( v = 2.0 xx 10^(16)Hz)`. The `v_(0)` Threshold frequency ) of the metal is

Options

A. `1.2xx10^(14)`Hz
B. `8xx10^(15)` Hz
C. `1.2xx10^(16)` Hz
D. `4xx10^(12)` Hz

Explanation

To solve the problem, we need to find the threshold frequency \( v_0 \) of the metal based on the given information about the kinetic energies of the emitted photoelectrons when the metal is irradiated with two different frequencies of light.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The kinetic energy (KE) of emitted photoelectrons can be expressed using the equation:
\[ KE = h \nu - h v_0 \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( v_0 \) is the threshold frequency.

2. **Setting Up the Equations**:
For the first frequency \( \nu_1 = 3.2 \times 10^{16} \, \text{Hz} \):
\[ KE_1 = h \nu_1 - h v_0 \]
For the second frequency \( \nu_2 = 2.0 \times 10^{16} \, \text{Hz} \):
\[ KE_2 = h \nu_2 - h v_0 \]

3. **Relating the Kinetic Energies**:
According to the problem, the kinetic energy of the photoelectrons emitted at frequency \( \nu_1 \) is twice that of those emitted at frequency \( \nu_2 \):
\[ KE_1 = 2 \times KE_2 \]

4. **Substituting the Kinetic Energy Equations**:
Substitute the expressions for \( KE_1 \) and \( KE_2 \):
\[ h \nu_1 - h v_0 = 2(h \nu_2 - h v_0) \]

5. **Simplifying the Equation**:
Expanding the equation gives:
\[ h \nu_1 - h v_0 = 2h \nu_2 - 2h v_0 \]
Rearranging terms leads to:
\[ h \nu_1 + h v_0 = 2h \nu_2 \]

6. **Isolating the Threshold Frequency**:
Dividing through by \( h \) (assuming \( h \neq 0 \)):
\[ \nu_1 + v_0 = 2 \nu_2 \]
Rearranging gives:
\[ v_0 = 2 \nu_2 - \nu_1 \]

7. **Substituting the Values**:
Now substitute \( \nu_1 = 3.2 \times 10^{16} \, \text{Hz} \) and \( \nu_2 = 2.0 \times 10^{16} \, \text{Hz} \):
\[ v_0 = 2(2.0 \times 10^{16}) - 3.2 \times 10^{16} \]
\[ v_0 = 4.0 \times 10^{16} - 3.2 \times 10^{16} \]
\[ v_0 = 0.8 \times 10^{16} \, \text{Hz} = 8.0 \times 10^{15} \, \text{Hz} \]

8. **Final Result**:
Thus, the threshold frequency \( v_0 \) of the metal is:
\[ \boxed{8.0 \times 10^{15} \, \text{Hz}} \]