The de Broglie wavelength of an electron `(m=9.11 xx 10^(-31) kg)` is `1.2 xx 10^(-10)m`. Determine the kinetic energy of the electron.

Options

A. `1.5 xx 10^(-15)J`
B. `1.7 xx 10^(-17)J`
C. `1.6 xx 10^(-16)J`
D. ` 1.8 xx 10^(-15)J`

Answer

To determine the kinetic energy of an electron given its de Broglie wavelength, we can follow these steps:

### Step 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:
- \( \lambda \) is the de Broglie wavelength,
- \( h \) is the Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \( m \) is the mass of the particle (for an electron, \(9.11 \times 10^{-31} \, \text{kg}\)),
- \( v \) is the velocity of the particle.

### Step 2: Rearranging the formula to find mv
From the de Broglie wavelength formula, we can rearrange it to find \( mv \):
\[ mv = \frac{h}{\lambda} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ mv = \frac{6.63 \times 10^{-34} \, \text{Js}}{1.2 \times 10^{-10} \, \text{m}} \]

### Step 4: Calculate mv
Calculating \( mv \):
\[ mv = \frac{6.63 \times 10^{-34}}{1.2 \times 10^{-10}} = 5.525 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the kinetic energy
The kinetic energy (KE) of the electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
We can express \( v \) in terms of \( mv \):
\[ KE = \frac{1}{2m} (mv)^2 \]
Substituting \( mv \) from the previous step:
\[ KE = \frac{1}{2m} \left(5.525 \times 10^{-24}\right)^2 \]

### Step 6: Substitute the mass of the electron
Now, substitute \( m = 9.11 \times 10^{-31} \, \text{kg} \):
\[ KE = \frac{1}{2 \times 9.11 \times 10^{-31}} \left(5.525 \times 10^{-24}\right)^2 \]

### Step 7: Calculate the kinetic energy
Calculating the value:
\[ KE = \frac{1}{2 \times 9.11 \times 10^{-31}} \times 3.055 \times 10^{-47} \]
\[ KE = \frac{3.055 \times 10^{-47}}{1.822 \times 10^{-30}} \approx 1.68 \times 10^{-17} \, \text{J} \]

### Final Answer
Thus, the kinetic energy of the electron is approximately:
\[ KE \approx 1.7 \times 10^{-17} \, \text{J} \]

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