Question 123: Atoms & Nuclei

Question

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Accepted Answer

The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:
$ E_n = -\frac{13.6 \, 	ext{eV}}{n^2} $
where:
$E_n$ is the energy in electron volts (eV),
$13.6 \, 	ext{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$),
$n$ is the principal quantum number (an integer starting from 1).
The Balmer Series
The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n \geq 3$ (i.e., from higher energy states) down to $n = 2$. These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.
Ground State Energy ($E_1$):
The energy of the electron in the ground state ($n=1$) of hydrogen is:
$ E_1 = -\frac{13.6 \, 	ext{eV}}{1^2} = -13.6 \, 	ext{eV} $
This represents the electron's energy level when it's closest to the nucleus.
Energy for the $n=3$ Level ($E_3$):
For an electron in the $n=3$ level, its energy is:
$ E_3 = -\frac{13.6 \, 	ext{eV}}{3^2} = -1.51 \, 	ext{eV} $
Minimum Energy for Balmer Series Transition:
The minimum energy transition within the Balmer series is from $n=3$ to $n=2$, but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$.
Energy Difference ($\Delta E$):
The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:
$ \Delta E = E_3 - E_1 = -1.51 \, 	ext{eV} - (-13.6 \, 	ext{eV}) = 12.09 \, 	ext{eV} $
Note
This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ($n=1$) to an excited state ($n=3$) from which it can then transition to $n=2$, emitting radiation observable as part of the Balmer series.

Answer

$13.6 \mathrm{eV}$

Answer

$1.5 \mathrm{eV}$

Answer

$12.1 \mathrm{eV}$

Answer

$1.9 \mathrm{eV}$

Atoms & Nuclei

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

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