When a hydrogen atom is excited from ground state to first excited state, then

Options

A. its kinetic energy increases by 10.2 eV
B. its kinetic energy decreases by 10.2 eV
C. its potential energy increases by 20.4 eV
D. its angular momentum increases by `1.05 xx 10^(-34) J-s`

Explanation

To solve the problem of what happens when a hydrogen atom is excited from the ground state to the first excited state, we can follow these steps:

### Step 1: Understand the Energy Levels
The energy levels of a hydrogen atom are given by the formula:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

where \( n \) is the principal quantum number. For the ground state (\( n=1 \)) and the first excited state (\( n=2 \)), we calculate the energies:

- Ground state (\( n=1 \)):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

- First excited state (\( n=2 \)):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \]

### Step 2: Calculate the Change in Energy
To find the change in energy when the atom is excited from the ground state to the first excited state, we subtract the initial energy from the final energy:

\[ \Delta E = E_2 - E_1 = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

### Step 3: Analyze Kinetic and Potential Energy
In a hydrogen atom, the total energy \( E \) is related to kinetic energy \( K \) and potential energy \( U \):

\[ E = K + U \]

From the relationships:
- \( K = -\frac{E}{2} \)
- \( U = 2E \)

For the ground state:
- Total energy \( E_1 = -13.6 \, \text{eV} \)
- Kinetic energy \( K_1 = -\frac{-13.6}{2} = 6.8 \, \text{eV} \)
- Potential energy \( U_1 = 2 \times -13.6 = -27.2 \, \text{eV} \)

For the first excited state:
- Total energy \( E_2 = -3.4 \, \text{eV} \)
- Kinetic energy \( K_2 = -\frac{-3.4}{2} = 1.7 \, \text{eV} \)
- Potential energy \( U_2 = 2 \times -3.4 = -6.8 \, \text{eV} \)

### Step 4: Determine Changes
Now we can analyze the changes in kinetic and potential energy:

- Change in kinetic energy:
\[ \Delta K = K_2 - K_1 = 1.7 \, \text{eV} - 6.8 \, \text{eV} = -5.1 \, \text{eV} \]

- Change in potential energy:
\[ \Delta U = U_2 - U_1 = -6.8 \, \text{eV} - (-27.2 \, \text{eV}) = 20.4 \, \text{eV} \]

### Conclusion
When a hydrogen atom is excited from the ground state to the first excited state:
- The total energy increases by \( 10.2 \, \text{eV} \).
- The kinetic energy decreases by \( 5.1 \, \text{eV} \).
- The potential energy increases by \( 20.4 \, \text{eV} \).