A metal is irradiated with light of frequency `3.2xx10^(16)Hz`. The ejected photoelectron has kinetic energy `(3)/(4)`the of energy of absorbed photon. The threshold frequency orf the metal would be :

Options

A. `1.6xx10^(15)Hz`
B. `8xx10^(15)Hz`
C. `2.4xx10^(25)Hz`
D. `2.4xx10^(18)Hz`

Explanation

To solve the problem, we need to find the threshold frequency (ν₀) of the metal when it is irradiated with light of a given frequency (ν) and the kinetic energy of the ejected photoelectron is given in relation to the energy of the absorbed photon.

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency of the light (ν) = \(3.2 \times 10^{16} \, \text{Hz}\)
- Kinetic energy of the ejected photoelectron = \(\frac{3}{4}\) of the energy of the absorbed photon.

2. **Recall the photoelectric equation:**
The energy of the absorbed photon can be expressed as:
\[ E = h \nu \]
where \(h\) is Planck's constant.

3. **Express the kinetic energy of the photoelectron:**
The kinetic energy (KE) of the ejected photoelectron can be expressed as:
\[ KE = \frac{3}{4} E = \frac{3}{4} h \nu \]

4. **Use the photoelectric equation:**
According to the photoelectric effect, the energy of the absorbed photon is equal to the sum of the threshold energy (E₀) and the kinetic energy (KE) of the photoelectron:
\[ E = E_0 + KE \]
Substituting the expressions we have:
\[ h \nu = E_0 + \frac{3}{4} h \nu \]

5. **Rearranging the equation:**
Rearranging gives:
\[ E_0 = h \nu - \frac{3}{4} h \nu = \frac{1}{4} h \nu \]

6. **Express the threshold frequency:**
The threshold energy can also be expressed as:
\[ E_0 = h \nu_0 \]
Therefore, we have:
\[ h \nu_0 = \frac{1}{4} h \nu \]
Dividing both sides by \(h\) gives:
\[ \nu_0 = \frac{1}{4} \nu \]

7. **Substituting the value of ν:**
Now substituting the value of ν:
\[ \nu_0 = \frac{1}{4} \times 3.2 \times 10^{16} \, \text{Hz} = \frac{3.2}{4} \times 10^{16} \, \text{Hz} = 0.8 \times 10^{16} \, \text{Hz} = 8 \times 10^{15} \, \text{Hz} \]

8. **Final Result:**
Thus, the threshold frequency (ν₀) of the metal is:
\[ \nu_0 = 8 \times 10^{15} \, \text{Hz} \]