The total energy of an electron in the first excited state of hydrogen atom is about `-3.4eV`. Its kinetic energy in this state is

Options

A. `-3.4eV`
B. `3.4eV`
C. `6.8eV`
D. `-6.8eV`

Answer

To find the kinetic energy of an electron in the first excited state of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between total energy, kinetic energy, and potential energy**:
The total energy (E) of an electron in a hydrogen atom is given by the equation:
\[ E = KE + PE \]
where \( KE \) is the kinetic energy and \( PE \) is the potential energy.

2. **Use the known relationship between kinetic energy and potential energy**:
In a hydrogen atom, the potential energy (PE) is related to the kinetic energy (KE) by the equation:
\[ PE = -2 \times KE \]
This means that the potential energy is always negative and twice the magnitude of the kinetic energy.

3. **Substitute the potential energy into the total energy equation**:
We can rewrite the total energy equation using the relationship between kinetic and potential energy:
\[ E = KE + PE = KE - 2 \times KE = -KE \]

4. **Substitute the given total energy value**:
We know from the problem that the total energy \( E \) in the first excited state is:
\[ E = -3.4 \, eV \]
Therefore, we can set up the equation:
\[ -KE = -3.4 \, eV \]

5. **Solve for kinetic energy**:
By multiplying both sides by -1, we find:
\[ KE = 3.4 \, eV \]

### Final Answer:
The kinetic energy of the electron in the first excited state of the hydrogen atom is \( 3.4 \, eV \).

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