De-broglie wavelength of a moving e– is decreased from 1Å to 0.5 Å, then calculate change in its K.E. in eV :

Options

A. 600 eV
B. 450 eV
C. 150 eV
D. 500 eV

Explanation

To solve the problem of calculating the change in kinetic energy of an electron when its De Broglie wavelength decreases from 1 Å to 0.5 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the De Broglie Wavelength Formula**:
The De Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \)
- \( m \) is the mass of the electron \( (9.11 \times 10^{-31} \, \text{kg}) \)
- \( v \) is the velocity of the electron.

2. **Relate Wavelength to Kinetic Energy**:
The kinetic energy (K.E.) of the electron can be expressed in terms of its velocity:
\[ K.E. = \frac{1}{2} mv^2 \]
We can also express \( v \) in terms of \( \lambda \):
\[ v = \frac{h}{m\lambda} \]

3. **Calculate Initial Kinetic Energy (K.E.1)**:
For the initial wavelength \( \lambda_1 = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \):
\[ v_1 = \frac{h}{m\lambda_1} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 1 \times 10^{-10}} \approx 7.27 \times 10^6 \, \text{m/s} \]
Now calculate the kinetic energy:
\[ K.E.1 = \frac{1}{2} mv_1^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (7.27 \times 10^6)^2 \approx 2.42 \times 10^{-14} \, \text{J} \]
Convert this to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ K.E.1 \approx \frac{2.42 \times 10^{-14}}{1.6 \times 10^{-19}} \approx 151.25 \, \text{eV} \]

4. **Calculate Final Kinetic Energy (K.E.2)**:
For the final wavelength \( \lambda_2 = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \):
\[ v_2 = \frac{h}{m\lambda_2} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 0.5 \times 10^{-10}} \approx 1.454 \times 10^7 \, \text{m/s} \]
Now calculate the kinetic energy:
\[ K.E.2 = \frac{1}{2} mv_2^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (1.454 \times 10^7)^2 \approx 6.93 \times 10^{-14} \, \text{J} \]
Convert this to electron volts:
\[ K.E.2 \approx \frac{6.93 \times 10^{-14}}{1.6 \times 10^{-19}} \approx 433.75 \, \text{eV} \]

5. **Calculate the Change in Kinetic Energy**:
\[ \Delta K.E. = K.E.2 - K.E.1 = 433.75 \, \text{eV} - 151.25 \, \text{eV} = 282.5 \, \text{eV} \]

### Final Answer:
The change in kinetic energy of the electron is approximately **282.5 eV**.