A hydrogen atom in ground state is given an energy of $10.2 \, \text{eV}$ . How many spectral lines will be emitted due to the transition of electrons?
A
$6$
B
$3$
C
$10$
D
$1$

Question

A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to the transition of electrons?

Accepted Answer

Given: A hydrogen atom is initially in the ground state and is given energy $10.2 \, \text{eV}$ .

Find: The number of spectral lines emitted due to the subsequent electronic transition.

For hydrogen atom, energy of the $n^{\text{th}}$ orbit varies as

E \propto \frac{Z^2}{n^2}

The energy of the second orbit is

E_2 = \frac{-13.6}{4} = -3.4 \, \text{eV}

Hence, the energy difference for the transition $2 \to 1$ is

\Delta E_{2 \to 1} = 10.2 \, \text{eV}

So, the given energy excites the electron only from $n = 1$ to $n = 2$ . After excitation, only one downward transition is possible:

n = 2 \to n = 1

Therefore, only one spectral line is emitted. The correct option is D.

Given: The electron starts from the ground state $n = 1$ and absorbs exactly $10.2 \, \text{eV}$ .

Find: How many emission lines can appear when it returns to a lower level.

The energy gap between the ground state and the first excited state of hydrogen is approximately $10.2 \, \text{eV}$ . Therefore, the electron is promoted from $n = 1$ to $n = 2$ .

Once the electron reaches $n = 2$ , it has only one lower energy level available, namely $n = 1$ . Thus only one radiative transition can occur:

n = 2 \rightarrow n = 1

Therefore, the number of spectral lines emitted is $1$ , so the correct option is D.

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