When a certain metal was irradiated with light of frequency `1.6xx10^(16)Hz`, the photoelectrons emitted had twice the kinetic energy as photoelectrons emitted when the same metal was irradiated with light of frequency `1.0xx10^(6)Hz`. Calculate the threshold frequency `(v_(0))` for the metal.

Explanation

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

### Step-by-Step Solution:

1. **Identify the given frequencies and their relationship to kinetic energy:**
- Let \( v_1 = 1.6 \times 10^{16} \) Hz (frequency of the first light).
- Let \( v_2 = 1.0 \times 10^{16} \) Hz (frequency of the second light).
- The kinetic energy of the photoelectrons emitted at \( v_1 \) is twice that at \( v_2 \):
\[ KE_1 = 2 \times KE_2 \]

2. **Use the photoelectric equation:**
The kinetic energy of the emitted photoelectrons can be expressed using the photoelectric effect 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.

3. **Write the equations for kinetic energy at both frequencies:**
- For frequency \( v_1 \):
\[ KE_1 = h v_1 - h v_0 \]
- For frequency \( v_2 \):
\[ KE_2 = h v_2 - h v_0 \]

4. **Substitute the relationship between kinetic energies:**
Since \( KE_1 = 2 \times KE_2 \), we can write:
\[ h v_1 - h v_0 = 2(h v_2 - h v_0) \]

5. **Simplify the equation:**
Expanding the right side:
\[ h v_1 - h v_0 = 2h v_2 - 2h v_0 \]
Rearranging gives:
\[ h v_1 - 2h v_2 = -h v_0 + 2h v_0 \]
\[ h v_1 - 2h v_2 = h v_0 \]

6. **Divide through by \( h \):**
\[ v_1 - 2 v_2 = v_0 \]

7. **Substitute the values of \( v_1 \) and \( v_2 \):**
\[ v_0 = 1.6 \times 10^{16} - 2 \times (1.0 \times 10^{16}) \]
\[ v_0 = 1.6 \times 10^{16} - 2.0 \times 10^{16} \]
\[ v_0 = -0.4 \times 10^{16} \]

8. **Correct the sign and express the threshold frequency:**
Since frequency cannot be negative, we take the absolute value:
\[ v_0 = 0.4 \times 10^{16} \text{ Hz} = 4.0 \times 10^{15} \text{ Hz} \]

### Final Answer:
The threshold frequency \( v_0 \) for the metal is:
\[ \boxed{4.0 \times 10^{15} \text{ Hz}} \]