The threshold frequency `v_@` for a metal is `0.5` x `10^(14)` s^(-1). What will be the kinetic energy of a photoelectron emitted when radiation of frequency v=`1.5` x `10^(15)` s^(-1) strikes on a metal surface?

Options

A. `h*10^(14)` J
B. `h*10^(16)` J
C. `h*10^(15) `J
D. `h J`

Answer

To solve the problem, we will use the photoelectric effect equation, which relates the kinetic energy of emitted photoelectrons to the frequency of the incident radiation and the threshold frequency of the metal.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Threshold frequency, \( v_0 = 0.5 \times 10^{14} \, \text{s}^{-1} \)
- Frequency of incident radiation, \( v = 1.5 \times 10^{15} \, \text{s}^{-1} \)

2. **Use the Photoelectric Effect Equation:**
The kinetic energy (KE) of the emitted photoelectron can be calculated using the formula:
\[ KE = h v - h v_0 \]
where \( h \) is Planck's constant.

3. **Factor out Planck's Constant:**
We can simplify the equation:
\[ KE = h (v - v_0) \]

4. **Substitute the Values:**
Now substitute the values of \( v \) and \( v_0 \):
\[ KE = h \left( 1.5 \times 10^{15} - 0.5 \times 10^{14} \right) \]

5. **Calculate the Difference in Frequencies:**
Convert \( 0.5 \times 10^{14} \) to the same power of ten as \( 1.5 \times 10^{15} \):
\[ 0.5 \times 10^{14} = 0.05 \times 10^{15} \]
Now, calculate:
\[ v - v_0 = 1.5 \times 10^{15} - 0.05 \times 10^{15} = 1.45 \times 10^{15} \]

6. **Final Expression for Kinetic Energy:**
Substitute back into the kinetic energy equation:
\[ KE = h (1.45 \times 10^{15}) \]

7. **Conclusion:**
The kinetic energy of the emitted photoelectron is:
\[ KE = 1.45 h \times 10^{15} \, \text{J} \]