Ratio between potential energy, kinetic energy and total energy of electron in hydrogen atom are

Options

A. `(KE)/( PE ) = - ( 1)/( 2)`
B. `( E )/( PE ) = ( 1)/( 2)`
C. `( KE )/( E ) = - ( 1)/( 2)`
D. `( E )/( PE + KE ) = - 1 `

Answer

To find the ratio between the potential energy (PE), kinetic energy (KE), and total energy (E) of an electron in a hydrogen atom, we can follow these steps:

### Step 1: Define the Kinetic Energy (KE)
The kinetic energy (KE) of an electron in a hydrogen atom is given by the formula:
\[ KE = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Define the Potential Energy (PE)
The potential energy (PE) of an electron in a hydrogen atom is related to the kinetic energy. It is given by:
\[ PE = -2 \times KE \]
Substituting the expression for KE:
\[ PE = -2 \left(-\frac{13.6 \, \text{eV}}{n^2}\right) = \frac{27.2 \, \text{eV}}{n^2} \]

### Step 3: Calculate the Total Energy (E)
The total energy (E) of the electron is the sum of the kinetic and potential energies:
\[ E = KE + PE \]
Substituting the expressions for KE and PE:
\[ E = -\frac{13.6 \, \text{eV}}{n^2} + \frac{27.2 \, \text{eV}}{n^2} = \frac{13.6 \, \text{eV}}{n^2} \]

### Step 4: Find the Ratios
Now we can find the ratios of KE, PE, and E:

1. **Ratio of KE to PE:**
\[ \frac{KE}{PE} = \frac{-\frac{13.6 \, \text{eV}}{n^2}}{\frac{27.2 \, \text{eV}}{n^2}} = -\frac{1}{2} \]

2. **Ratio of E to PE:**
\[ \frac{E}{PE} = \frac{\frac{13.6 \, \text{eV}}{n^2}}{\frac{27.2 \, \text{eV}}{n^2}} = \frac{1}{2} \]

3. **Ratio of KE to E:**
\[ \frac{KE}{E} = \frac{-\frac{13.6 \, \text{eV}}{n^2}}{\frac{13.6 \, \text{eV}}{n^2}} = -1 \]

4. **Ratio of E to (PE + KE):**
Since \( E = PE + KE \), we have:
\[ \frac{E}{PE + KE} = \frac{E}{E} = 1 \]

### Summary of Ratios
- \( \frac{KE}{PE} = -\frac{1}{2} \)
- \( \frac{E}{PE} = \frac{1}{2} \)
- \( \frac{KE}{E} = -1 \)
- \( \frac{E}{PE + KE} = 1 \)