If a light with frequency `4xx10^(16) Hz` emitted photoelectrons with double the maximum kinetic as are emitted by the light of frequency `2.5xx10^(16) Hz` from the same metal surface, then what is the threshold frequency `(v_(0))` of the metal?

Explanation

To solve the problem, we will use the photoelectric effect equations and the information provided about the frequencies and kinetic energies of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of the incident photons is given by the equation:
\[ E = h \nu \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(\nu\) is the frequency of the light.

2. **Setting Up the Equations**:
For the first frequency \( \nu_1 = 4 \times 10^{16} \, \text{Hz} \):
\[ h \nu_1 = h \nu_0 + KE_1 \]
where \(KE_1\) is the kinetic energy of the photoelectrons emitted by this frequency and \(\nu_0\) is the threshold frequency.

For the second frequency \( \nu_2 = 2.5 \times 10^{16} \, \text{Hz} \):
\[ h \nu_2 = h \nu_0 + KE_2 \]
where \(KE_2\) is the kinetic energy of the photoelectrons emitted by this frequency.

3. **Relating Kinetic Energies**:
According to the problem, the kinetic energy of the photoelectrons emitted by the first frequency is double that of the second frequency:
\[ KE_1 = 2 KE_2 \]

4. **Substituting Kinetic Energies**:
From the first equation:
\[ KE_1 = h \nu_1 - h \nu_0 \]
From the second equation:
\[ KE_2 = h \nu_2 - h \nu_0 \]
Substituting \(KE_2\) into the relation \(KE_1 = 2 KE_2\):
\[ h \nu_1 - h \nu_0 = 2(h \nu_2 - h \nu_0) \]

5. **Expanding and Rearranging**:
Expanding the equation gives:
\[ h \nu_1 - h \nu_0 = 2h \nu_2 - 2h \nu_0 \]
Rearranging terms leads to:
\[ h \nu_1 + h \nu_0 = 2h \nu_2 \]

6. **Dividing by \(h\)**:
Dividing the entire equation by \(h\) (assuming \(h \neq 0\)):
\[ \nu_1 + \nu_0 = 2 \nu_2 \]

7. **Substituting the Frequencies**:
Now substituting the known values:
\[ 4 \times 10^{16} + \nu_0 = 2(2.5 \times 10^{16}) \]
Simplifying the right side:
\[ 4 \times 10^{16} + \nu_0 = 5 \times 10^{16} \]

8. **Solving for the Threshold Frequency**:
Rearranging gives:
\[ \nu_0 = 5 \times 10^{16} - 4 \times 10^{16} \]
\[ \nu_0 = 1 \times 10^{16} \, \text{Hz} \]

### Final Answer:
The threshold frequency \( \nu_0 \) of the metal is:
\[ \nu_0 = 1 \times 10^{16} \, \text{Hz} \]