The threshold frequency `(v_(0))` for a metal is `5 xx 10^(14)S^(-1)`. Calculate the kinetic energy of emitted electron when a radiation with frequency of `2 xx 10^(15) sec^(-1)` hits the metal surface.

Answer

To solve the problem of calculating the kinetic energy of an emitted electron when radiation with a frequency of \(2 \times 10^{15} \, \text{s}^{-1}\) hits a metal surface with a threshold frequency of \(5 \times 10^{14} \, \text{s}^{-1}\), we can follow these steps:

### Step 1: Understand the concept of threshold frequency
The threshold frequency (\(v_0\)) is the minimum frequency of radiation required to eject an electron from the surface of a metal. If the frequency of the incoming radiation (\(v\)) is greater than the threshold frequency, the excess energy is converted into kinetic energy of the emitted electron.

### Step 2: Write the formula for kinetic energy
The kinetic energy (KE) of the emitted electron can be calculated using the formula:
\[ KE = h \cdot v - h \cdot v_0 \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(v\) is the frequency of the incoming radiation,
- \(v_0\) is the threshold frequency.

### Step 3: Substitute the values into the formula
Given:
- \(v = 2 \times 10^{15} \, \text{s}^{-1}\)
- \(v_0 = 5 \times 10^{14} \, \text{s}^{-1}\)

Substituting these values into the formula:
\[ KE = h \cdot (2 \times 10^{15}) - h \cdot (5 \times 10^{14}) \]
\[ KE = h \cdot \left(2 \times 10^{15} - 5 \times 10^{14}\right) \]
\[ KE = h \cdot (1.5 \times 10^{15}) \]

### Step 4: Calculate the kinetic energy
Now, substituting the value of \(h\):
\[ KE = (6.626 \times 10^{-34}) \cdot (1.5 \times 10^{15}) \]
\[ KE = 9.939 \times 10^{-19} \, \text{J} \]

### Step 5: Final result
Thus, the kinetic energy of the emitted electron is approximately:
\[ KE \approx 9.94 \times 10^{-19} \, \text{J} \]