Question

The ionisation potential of hydrogen atom is 13.6 eV. How much energy need to be supplied to ionise the hydrogen atom in the first excited state?

• A. 13.6 eV
• B. 27.2 eV
• C. 3.4 eV
• D. 6.8 eV

Hint:

Ionisation energy = $|Eₙ|$ for that state.

Answer 1

The Correct Option is C

The ionisation potential of a hydrogen atom is given as 13.6 eV. This is the energy required to remove an electron from the ground state (n=1) of a hydrogen atom. To solve this problem, we need to determine the energy required to ionise the hydrogen atom from its first excited state (n=2).

1. The energy levels of a hydrogen atom are given by the formula:

\(E_n = -\frac{13.6}{n^2} \, \text{eV}\)

1. For the first excited state (n=2), the energy level is:

\(E_2 = -\frac{13.6}{2^2} \, \text{eV} = -\frac{13.6}{4} \, \text{eV} = -3.4 \, \text{eV}\)

1. To ionise the hydrogen atom from this excited state, the electron must be provided enough energy to reach 0 eV (which is the level at which the electron can escape the atom, i.e., ionisation).

Thus, the energy required = 0 eV - (-3.4 eV) = 3.4 eV.

1. Therefore, the amount of energy needed to ionise the hydrogen atom in its first excited state is 3.4 eV. Hence, the correct answer is option:

3.4 eV

1. .

Other options can be ruled out as follows:

• \(13.6 \, \text{eV}\) is the energy required from the ground state (not applicable here as we are considering the first excited state).
• \(27.2 \, \text{eV}\) is irrelevant to our calculations involving n=2.
• \(6.8 \, \text{eV}\) does not match the calculation from energy level \(n=2\).