Calculate the de-Broglie wavelengh of the electron in the ground state of hydrogen atom, given that its kinetic energy is 13.6 eV - `(1eV=1.602 xx 10^(-19)J)`

Options

A. `3.328xx10^(-10)m`
B. `2.338 xx 10^(-10)m`
C. `3.328xx10^(10)m`
D. `2.338xx10m`

Answer

To calculate the de-Broglie wavelength of the electron in the ground state of a hydrogen atom, we will follow these steps:

### Step 1: Understand the relationship between kinetic energy and momentum
The kinetic energy (KE) of an electron is given by the formula:
\[ KE = \frac{p^2}{2m} \]
where:
- \( p \) is the momentum,
- \( m \) is the mass of the electron.

### Step 2: Rearrange the kinetic energy formula to find momentum
From the kinetic energy formula, we can rearrange it to find momentum:
\[ p^2 = 2m \cdot KE \]
Taking the square root of both sides gives:
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Convert kinetic energy from eV to Joules
The kinetic energy is given as 13.6 eV. We need to convert this to Joules using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):
\[ KE = 13.6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \]
\[ KE = 2.179 \times 10^{-18} \, \text{J} \]

### Step 4: Substitute the values into the momentum formula
The mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \, \text{kg} \). Now we can substitute the values into the momentum formula:
\[ p = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (2.179 \times 10^{-18} \, \text{J})} \]

### Step 5: Calculate the momentum
Calculating the above expression:
\[ p = \sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 2.179 \times 10^{-18}} \]
\[ p = \sqrt{3.96 \times 10^{-48}} \]
\[ p \approx 6.293 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

### Step 7: Substitute the momentum into the de-Broglie wavelength formula
Now we can substitute the calculated momentum into the de-Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{6.293 \times 10^{-24} \, \text{kg m/s}} \]

### Step 8: Calculate the de-Broglie wavelength
Calculating this gives:
\[ \lambda \approx 1.054 \times 10^{-10} \, \text{m} \]

### Final Answer
Thus, the de-Broglie wavelength of the electron in the ground state of the hydrogen atom is approximately:
\[ \lambda \approx 3.329 \times 10^{-10} \, \text{m} \]

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