The kinetic energy of electron is `3.0 xx 10^(-25)J`. The wave length of the electron is

Options

A. `7965A^(0)`
B. `4625A^(0)`
C. `91A^(0)`
D. `8967A^(0)`

Explanation

To find the wavelength of the electron given its kinetic energy, we can use the de Broglie wavelength formula, which is given by:

\[ \lambda = \frac{h}{mv} \]

where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the velocity of the electron.

### Step 1: Calculate the velocity of the electron using kinetic energy

The kinetic energy (KE) of the electron is given by:

\[ KE = \frac{1}{2} mv^2 \]

From this, we can express the velocity \(v\) as:

\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

Substituting the given kinetic energy \(KE = 3.0 \times 10^{-25} \, \text{J}\) and the mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\):

\[ v = \sqrt{\frac{2 \cdot 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}} \]

### Step 2: Substitute the velocity back into the wavelength formula

Now, substitute the expression for \(v\) back into the de Broglie wavelength formula:

\[ \lambda = \frac{h}{\sqrt{\frac{2 \cdot KE}{m}}} \]

This can be rewritten as:

\[ \lambda = \frac{h}{\sqrt{2 \cdot KE}} \cdot \sqrt{m} \]

### Step 3: Substitute the values into the wavelength formula

Now, substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 3.0 \times 10^{-25}}} \cdot \sqrt{9.1 \times 10^{-31}} \]

### Step 4: Calculate the wavelength

First, calculate the denominator:

\[ \sqrt{2 \cdot 3.0 \times 10^{-25}} = \sqrt{6.0 \times 10^{-25}} = 7.746 \times 10^{-13} \]

Now, calculate \(\sqrt{9.1 \times 10^{-31}}\):

\[ \sqrt{9.1 \times 10^{-31}} = 3.016 \times 10^{-16} \]

Now, substituting these values back into the wavelength equation:

\[ \lambda = \frac{6.626 \times 10^{-34}}{7.746 \times 10^{-13}} \cdot 3.016 \times 10^{-16} \]

Calculating this gives:

\[ \lambda \approx 0.8967 \times 10^{-6} \, \text{m} \]

### Step 5: Convert to Angstroms

To convert meters to Angstroms (1 Angstrom = \(10^{-10}\) m):

\[ \lambda \approx 0.8967 \times 10^{-6} \, \text{m} = 8.967 \times 10^{-7} \, \text{m} = 8967 \, \text{Å} \]

### Final Answer

The wavelength of the electron is approximately \(8967 \, \text{Å}\).

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