A radiation of wave length 3000 Å is required to remove an electron from a metal atom. If a radiation of wave length 2000Å is allowed to incident on the metal surface, the kinetic energy of the emitted electron in kJ/ mole is

Options

A. `1.98 xx 10^(2)`
B. `3.3 xx 10^(-19)`
C. `3.3 xx 10^(-20)`
D. 19.8

Answer

To solve the problem, we need to calculate the kinetic energy of the emitted electron when a radiation of wavelength 2000 Å is incident on a metal surface, given that the threshold wavelength required to remove an electron from the metal atom is 3000 Å.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Threshold wavelength (λ₀) = 3000 Å = 3000 × 10⁻¹⁰ m
- Incident wavelength (λ) = 2000 Å = 2000 × 10⁻¹⁰ m

2. **Convert Wavelengths to Meters:**
- λ₀ = 3000 Å = 3000 × 10⁻¹⁰ m
- λ = 2000 Å = 2000 × 10⁻¹⁰ m

3. **Calculate the Threshold Frequency (ν₀):**
The frequency can be calculated using the formula:
\[ \nu_0 = \frac{c}{\lambda_0} \]
where \(c\) is the speed of light (approximately \(3 \times 10^8\) m/s).
\[ \nu_0 = \frac{3 \times 10^8 \text{ m/s}}{3000 \times 10^{-10} \text{ m}} = 1 \times 10^{15} \text{ Hz} \]

4. **Calculate the Frequency of the Incident Radiation (ν):**
Similarly, we calculate the frequency for the incident wavelength:
\[ \nu = \frac{c}{\lambda} \]
\[ \nu = \frac{3 \times 10^8 \text{ m/s}}{2000 \times 10^{-10} \text{ m}} = 1.5 \times 10^{15} \text{ Hz} \]

5. **Calculate the Kinetic Energy (KE) of the Emitted Electron:**
The kinetic energy of the emitted electron can be calculated using the formula:
\[ KE = h\nu - h\nu_0 \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\)).
\[ KE = h(\nu - \nu_0) \]
\[ KE = 6.626 \times 10^{-34} \text{ J s} \times (1.5 \times 10^{15} \text{ Hz} - 1 \times 10^{15} \text{ Hz}) \]
\[ KE = 6.626 \times 10^{-34} \text{ J s} \times 0.5 \times 10^{15} \text{ Hz} \]
\[ KE = 3.313 \times 10^{-19} \text{ J} \]

6. **Convert Kinetic Energy to kJ/mol:**
To convert Joules to kilojoules per mole, we use the conversion factor \(1 \text{ mol} = 6.022 \times 10^{23} \text{ particles}\):
\[ KE = \frac{3.313 \times 10^{-19} \text{ J} \times 1000 \text{ J/kJ}}{6.022 \times 10^{23}} = 5.50 \times 10^{-14} \text{ kJ} \]
\[ KE \approx 0.0000550 \text{ kJ} \text{ per electron} \]
Now, multiplying by Avogadro's number:
\[ KE \approx 3.3 \text{ kJ/mol} \]

### Final Answer:
The kinetic energy of the emitted electron is approximately **3.3 kJ/mol**.