The threshold frequency `v_@` for a metal is `0.5` x `10^(15) 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 find the kinetic energy of the photoelectron emitted when radiation strikes a metal surface, we can follow these steps:

### Step 1: Understand the given values
- Threshold frequency, \( v_0 = 0.5 \times 10^{15} \, \text{s}^{-1} \)
- Frequency of the radiation, \( v = 1.5 \times 10^{15} \, \text{s}^{-1} \)

### Step 2: Recall the formula for kinetic energy of emitted photoelectrons
The kinetic energy (KE) of the emitted photoelectron can be calculated using the formula:
\[ KE = h \cdot v - h \cdot v_0 \]
Where:
- \( h \) is Planck's constant
- \( v \) is the frequency of the incident radiation
- \( v_0 \) is the threshold frequency

### Step 3: Factor out Planck's constant
We can factor out \( h \) from the equation:
\[ KE = h(v - v_0) \]

### Step 4: Substitute the values of \( v \) and \( v_0 \)
Now, substituting the values of \( v \) and \( v_0 \):
\[ KE = h \left(1.5 \times 10^{15} - 0.5 \times 10^{15}\right) \]

### Step 5: Simplify the expression
Calculating the difference:
\[ 1.5 \times 10^{15} - 0.5 \times 10^{15} = 1.0 \times 10^{15} \]
Thus, we have:
\[ KE = h \cdot 1.0 \times 10^{15} \]

### Step 6: Write the final expression for kinetic energy
The final expression for the kinetic energy of the emitted photoelectron is:
\[ KE = h \times 10^{15} \, \text{Joule} \]

### Conclusion
The kinetic energy of the photoelectron emitted when radiation of frequency \( v = 1.5 \times 10^{15} \, \text{s}^{-1} \) strikes the metal surface is:
\[ KE = h \times 10^{15} \, \text{Joule} \]