If an electron has an energy such that its de Broglie wavelength is `5500 Å`, then the energy value of that electron is `(h= 6.6 xx 10^(-34))` Js, `m_( c) = 9.1 xx 10^(-31)` kg

Options

A. `8xx10^(-20)J`
B. `8xx10^(-10)J`
C. 8J
D. `8xx10^(-25)J`

Answer

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

### Step 1: Write down the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where:
- \( \lambda \) is the de Broglie wavelength,
- \( h \) is Planck's constant,
- \( p \) is the momentum of the electron.

### Step 2: Express momentum in terms of kinetic energy
The momentum (\( p \)) of an electron can be expressed in terms of its kinetic energy (\( KE \)):
\[ p = \sqrt{2m \cdot KE} \]
where:
- \( m \) is the mass of the electron.

### Step 3: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Rearrange the equation to solve for kinetic energy
Squaring both sides of the equation:
\[ \lambda^2 = \frac{h^2}{2m \cdot KE} \]
Rearranging for kinetic energy gives:
\[ KE = \frac{h^2}{2m \cdot \lambda^2} \]

### Step 5: Substitute the known values
Now we can substitute the known values into the equation:
- \( h = 6.6 \times 10^{-34} \) Js,
- \( m = 9.1 \times 10^{-31} \) kg,
- \( \lambda = 5500 \) Å = \( 5500 \times 10^{-10} \) m = \( 5.5 \times 10^{-7} \) m.

Substituting these values:
\[ KE = \frac{(6.6 \times 10^{-34})^2}{2 \cdot (9.1 \times 10^{-31}) \cdot (5.5 \times 10^{-7})^2} \]

### Step 6: Calculate the kinetic energy
Calculating the numerator:
\[ (6.6 \times 10^{-34})^2 = 4.356 \times 10^{-67} \text{ J}^2 \]

Calculating the denominator:
\[ 2 \cdot (9.1 \times 10^{-31}) \cdot (5.5 \times 10^{-7})^2 = 2 \cdot (9.1 \times 10^{-31}) \cdot (3.025 \times 10^{-13}) = 5.5 \times 10^{-43} \text{ kg m}^2 \]

Now substituting back:
\[ KE = \frac{4.356 \times 10^{-67}}{5.5 \times 10^{-43}} \approx 7.91 \times 10^{-25} \text{ J} \]

### Step 7: Final answer
Thus, the energy value of the electron is approximately:
\[ KE \approx 8 \times 10^{-25} \text{ J} \]