In an experiment on photoelectric emission from a metallic surface ,wavelength of incident light `2xx10^(-7)` m and stopping potential is 2.5V .The threshold frequency of the metal (in Hz) approximately
(Charge of electrons `e=1.6xx10^(-19)`C, Planck's constant ,`h=6.6xx10^(-34)J-s)`

Options

A. `12xx10^(15)`
B. `9xx10^(15)`
C. `9xx10^(14)`
D. `12xx10^(13)`

Answer

To find the threshold frequency of the metal in the photoelectric emission experiment, we will use the relationship between the stopping potential, the energy of the incident photons, and the threshold frequency.

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength of incident light, \( \lambda = 2 \times 10^{-7} \) m
- Stopping potential, \( V = 2.5 \) V
- Charge of electron, \( e = 1.6 \times 10^{-19} \) C
- Planck's constant, \( h = 6.63 \times 10^{-34} \) J·s
- Speed of light, \( c = 3 \times 10^{8} \) m/s

2. **Calculate the energy of the incident photons:**
The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J·s}) \times (3 \times 10^{8} \, \text{m/s})}{2 \times 10^{-7} \, \text{m}} \]

3. **Perform the calculation:**
\[ E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{2 \times 10^{-7}} = \frac{1.989 \times 10^{-25}}{2 \times 10^{-7}} = 9.945 \times 10^{-19} \, \text{J} \]

4. **Relate stopping potential to the energy of emitted electrons:**
The stopping potential \( V \) relates to the kinetic energy of the emitted electrons:
\[ eV = E - E_0 \]
where \( E_0 \) is the work function of the metal. Rearranging gives:
\[ E_0 = E - eV \]

5. **Calculate the work function \( E_0 \):**
Substituting the values:
\[ E_0 = 9.945 \times 10^{-19} - (1.6 \times 10^{-19} \times 2.5) \]
\[ E_0 = 9.945 \times 10^{-19} - 4.0 \times 10^{-19} = 5.945 \times 10^{-19} \, \text{J} \]

6. **Find the threshold frequency \( f_0 \):**
The threshold frequency \( f_0 \) can be calculated using the work function:
\[ E_0 = h f_0 \implies f_0 = \frac{E_0}{h} \]
Substituting the values:
\[ f_0 = \frac{5.945 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 8.96 \times 10^{14} \, \text{Hz} \]

7. **Round to the nearest significant figure:**
The approximate threshold frequency is:
\[ f_0 \approx 9 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The threshold frequency of the metal is approximately \( 9 \times 10^{14} \, \text{Hz} \).

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