What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $-2.18 \times 10^{-11}$ ergs.

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to

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-2.18 \times 10^{-11}

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Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

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

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

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The electron must absorb $2.09 \times 10^{-18}$ J to jump from $n=1$ to $n=5$ ; when it falls back, it emits light of wavelength $9.50 \times 10^{-8}$ m (95.0 nm, in the ultraviolet).

The Bohr model tells us that electrons in hydrogen occupy discrete energy levels labeled by the principal quantum number $n = 1, 2, 3, \ldots$ . The energy of an electron in the $n$ -th orbit is given by

E_n = \frac{E_1}{n^2}

where $E_1$ is the ground-state energy (the most negative, lowest energy). Because $E_1$ is negative, higher orbits ( $n = 2, 3, \ldots$ ) have less negative (higher) energies—the electron is less tightly bound.

When an electron jumps from a lower orbit to a higher one, it must absorb energy equal to the difference $\Delta E = E_{\text{final}} - E_{\text{initial}}$ . When it falls back down, it emits a photon carrying exactly that energy, and the photon's wavelength is determined by $E = \frac{hc}{\lambda}$ .

Step-by-step solution

1. Convert the ground-state energy to SI units.

We are given $E_1 = -2.18 \times 10^{-11}$ erg. Since $1 \text{ erg} = 10^{-7}$ J,

E_1 = -2.18 \times 10^{-11} \times 10^{-7} = -2.18 \times 10^{-18} \text{ J}.

2. Find the energy of the fifth Bohr orbit.

Using the quantization formula,

E_5 = \frac{E_1}{5^2} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \text{ J}.

3. Calculate the energy required to excite the electron from $n=1$ to $n=5$ .

The energy absorbed is

\Delta E = E_5 - E_1 = \left(-8.72 \times 10^{-20}\right) - \left(-2.18 \times 10^{-18}\right).

Factor out the common power of ten:

\Delta E = -8.72 \times 10^{-20} + 2.18 \times 10^{-18} = -8.72 \times 10^{-20} + 218 \times 10^{-20} = 209.28 \times 10^{-20} \text{ J}.

Simplifying,

\Delta E = 2.09 \times 10^{-18} \text{ J}.

This is the energy the atom must absorb to promote the electron from the ground state to the fifth orbit.

\Delta E = E_1 \left(1 - \frac{1}{n^2}\right)

4. Find the wavelength of the photon emitted when the electron returns to the ground state.

When the electron falls from $n=5$ back to $n=1$ , it emits a photon with energy $2.09 \times 10^{-18}$ J. The photon energy and wavelength are related by

$E = \frac{hc}{\lambda} \quad \Rightarrow \quad \lambda = \frac{hc}{E}.$ …

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