Determine the kinetic energy of an electron that has a de Broglie wavelength equal to twice the diameter of the hydrogen atom. Assume that the hydrogen atom is a sphere of radius `5.3 xx 10^(-11)` m.

Options

A. 13.6 eV
B. 33.6 eV
C. 27.2 eV
D. 48.9 eV

Answer

To determine the kinetic energy of an electron that has a de Broglie wavelength equal to twice the diameter of a hydrogen atom, we can follow these steps:

### Step 1: Calculate the de Broglie wavelength
The diameter of the hydrogen atom can be calculated using its radius. Given that the radius of the hydrogen atom is \( r = 5.3 \times 10^{-11} \) m, the diameter \( d \) is:
\[ d = 2r = 2 \times 5.3 \times 10^{-11} \, \text{m} = 1.06 \times 10^{-10} \, \text{m} \]
Thus, the de Broglie wavelength \( \lambda \) is:
\[ \lambda = 2d = 2 \times 1.06 \times 10^{-10} \, \text{m} = 2.12 \times 10^{-10} \, \text{m} \]

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{mv} \]
Where:
- \( h \) is Planck's constant (\( 6.67 \times 10^{-34} \, \text{Js} \))
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \))
- \( v \) is the velocity of the electron.

From this, we can rearrange to find momentum \( p \):
\[ p = mv = \frac{h}{\lambda} \]

### Step 3: Calculate the momentum
Substituting the values:
\[ p = \frac{6.67 \times 10^{-34} \, \text{Js}}{2.12 \times 10^{-10} \, \text{m}} \approx 3.15 \times 10^{-24} \, \text{kg m/s} \]

### Step 4: Relate momentum to kinetic energy
The kinetic energy \( KE \) can be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
Substituting the momentum we calculated:
\[ KE = \frac{(3.15 \times 10^{-24})^2}{2 \times 9.1 \times 10^{-31}} \]

### Step 5: Calculate the kinetic energy
Calculating \( p^2 \):
\[ p^2 = (3.15 \times 10^{-24})^2 \approx 9.92 \times 10^{-48} \, \text{kg}^2 \text{m}^2/\text{s}^2 \]
Now substituting into the kinetic energy formula:
\[ KE = \frac{9.92 \times 10^{-48}}{2 \times 9.1 \times 10^{-31}} \approx \frac{9.92 \times 10^{-48}}{1.82 \times 10^{-30}} \approx 5.45 \times 10^{-18} \, \text{J} \]

### Step 6: Convert kinetic energy to electron volts
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE \approx \frac{5.45 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 34.06 \, \text{eV} \]

### Final Answer
Thus, the kinetic energy of the electron is approximately:
\[ \text{KE} \approx 34.06 \, \text{eV} \]