`K_("max")` (in eV) = 3, and frequency of light (in Hz) `=1xx10^(15)` for a metal used as a cathode in a photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is -

Options

A. `1xx10^(14)" Hz"`
B. `1.5xx10^(14)" Hz"`
C. `2.1xx10^(14)" Hz"`
D. `2.7xx10^(14)" Hz"`

Answer

To find the threshold frequency of light for the photoelectric emission from the metal, we can use the photoelectric equation:

\[ K_{\text{max}} = hf - hf_0 \]

Where:
- \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons.
- \( h \) is Planck's constant.
- \( f \) is the frequency of the incident light.
- \( f_0 \) is the threshold frequency.

We can rearrange this equation to solve for the threshold frequency \( f_0 \):

\[ f_0 = f - \frac{K_{\text{max}}}{h} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Maximum kinetic energy, \( K_{\text{max}} = 3 \, \text{eV} \)
- Frequency of light, \( f = 1 \times 10^{15} \, \text{Hz} \)
- Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J s} \)

2. **Convert the maximum kinetic energy from eV to Joules:**
- We know that \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).
- Therefore, \( K_{\text{max}} = 3 \, \text{eV} = 3 \times 1.6 \times 10^{-19} \, \text{J} = 4.8 \times 10^{-19} \, \text{J} \).

3. **Substitute the values into the equation for \( f_0 \):**
\[ f_0 = f - \frac{K_{\text{max}}}{h} \]
\[ f_0 = 1 \times 10^{15} - \frac{4.8 \times 10^{-19}}{6.626 \times 10^{-34}} \]

4. **Calculate \( \frac{K_{\text{max}}}{h} \):**
\[ \frac{K_{\text{max}}}{h} = \frac{4.8 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 7.24 \times 10^{14} \, \text{Hz} \]

5. **Now substitute this back into the equation for \( f_0 \):**
\[ f_0 = 1 \times 10^{15} - 7.24 \times 10^{14} = 0.276 \times 10^{15} \, \text{Hz} \]

6. **Convert \( f_0 \) into a more standard form:**
\[ f_0 \approx 2.76 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The threshold frequency \( f_0 \) for the photoelectric emission from the metal is approximately \( 2.76 \times 10^{14} \, \text{Hz} \).