If the potential energy of the electron in the first allowed orbit in hydrogen atom is `E`: its

Options

A. ionization potential is `E`: its
B. kinetic energy is `-E//2`
C. total energy is `E//2`
D. none of these

Explanation

To solve the problem regarding the potential energy of the electron in the first allowed orbit of a hydrogen atom, we need to understand the concept of potential energy in the context of atomic physics.

### Step-by-Step Solution:

1. **Understanding Potential Energy in Hydrogen Atom**:
The potential energy (U) of an electron in a hydrogen atom is given by the formula:
\[ U = -\frac{k e^2}{r} \]
where \( k \) is Coulomb's constant, \( e \) is the charge of the electron, and \( r \) is the radius of the orbit.

2. **Finding the Radius of the First Orbit**:
For the hydrogen atom, the radius of the first allowed orbit (Bohr radius) is:
\[ r_1 = \frac{4 \pi \epsilon_0 \hbar^2}{m e^2} \]
where \( \epsilon_0 \) is the permittivity of free space, \( \hbar \) is the reduced Planck's constant, and \( m \) is the mass of the electron.

3. **Substituting the Radius into the Potential Energy Formula**:
Now, substituting \( r_1 \) into the potential energy formula:
\[ U_1 = -\frac{k e^2}{r_1} \]

4. **Using Known Constants**:
We can express \( k \) in terms of \( \epsilon_0 \):
\[ k = \frac{1}{4 \pi \epsilon_0} \]
Therefore, the potential energy becomes:
\[ U_1 = -\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{r_1} \]

5. **Calculating the Potential Energy**:
By substituting \( r_1 \) into the equation, we can find the potential energy \( E \):
\[ E = -\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{\frac{4 \pi \epsilon_0 \hbar^2}{m e^2}} = -\frac{m e^4}{(4 \pi \epsilon_0)^2 \hbar^2} \]

6. **Conclusion**:
The potential energy of the electron in the first allowed orbit of the hydrogen atom is given by:
\[ E = -\frac{m e^4}{(4 \pi \epsilon_0)^2 \hbar^2} \]