Question 52: Atoms & Nuclei

Question

The ground state energy of hydrogen atom is $$-13.6 \mathrm{~eV}$$. If the electron jumps from the $$3^{	ext {rd }}$$ excited state to the ground state then the energy of the radiation emitted will be:

Accepted Answer

To determine the energy of the radiation emitted when an electron in hydrogen transitions from the $3^{	ext{rd}}$ excited state to the ground state, first, we need to understand what is meant by "ground state" and "excited states" in the context of a hydrogen atom.
The energy levels of an electron in a hydrogen atom are given by the formula:
$$E_n = -\frac{13.6 \ eV}{n^2}$$
where $E_n$ is the energy of the n-th energy level (with the ground state being $n = 1$) and 13.6 eV is the energy of the electron in the ground state. The ground state (n=1) has an energy of $-13.6$ eV. The $3^{	ext{rd}}$ excited state would correspond to $n = 4$, since the excited states start counting after the ground state $n=1$ (first excited state is $n=2$, second is $n=3$, and so forth).
Therefore, the energy of the $3^{	ext{rd}}$ excited state ($n=4$) is:
$$E_4 = -\frac{13.6 \ eV}{4^2} = -\frac{13.6 \ eV}{16} = -0.85 \ eV$$
To find the energy of the radiation emitted when the electron transitions from the $n = 4$ level to the ground state ($n = 1$), we calculate the difference in energy between these two levels:
$$\Delta E = E_{final} - E_{initial} = E_1 - E_4 = (-13.6 \ eV) - (-0.85 \ eV) = -13.6 \ eV + 0.85 \ eV = -12.75 \ eV$$
The negative sign indicates that energy is released, so the magnitude of the energy is 12.75 eV. Thus, the energy of the radiation emitted is
12.75 eV
.
This corresponds to
Option B
.

Answer

1.275 MeV

Answer

12.75 eV

Answer

12.75 J

Answer

12.75 MeV

Atoms & Nuclei

The ground state energy of hydrogen atom is $-13.6 \mathrm{~eV}. If the electron jumps from the 3^{\text {rd }}$ excited state to the ground state then the energy of the radiation emitted will be:

Choose the correct answer: