Bohr's model — calculate energy of electron in nth orbit of hydrogen

Question

Using Bohr’s model, derive the expression for the energy of an electron in the $n$ th orbit of hydrogen atom. Calculate the energy of the electron in the 3rd orbit. What wavelength of light is emitted when an electron falls from $n = 3$ to $n = 2$ ?

(JEE Main 2022, similar pattern)

Solution — Step by Step

In Bohr’s model, the total energy (KE + PE) of an electron in orbit $n$ is:

E_n = -\frac{me^4}{8\varepsilon_0^2 h^2} \cdot \frac{Z^2}{n^2} = -13.6 \times \frac{Z^2}{n^2} \text{ eV}

For hydrogen ( $Z = 1$ ):

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

The negative sign means the electron is bound — you need to supply $|E_n|$ to free it.

E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9}

\boxed{E_3 = -1.51 \text{ eV}}

Energy of emitted photon:

\Delta E = E_3 - E_2 = -1.51 - (-3.4) = 1.89 \text{ eV}

Using $E = hc/\lambda$ :

\lambda = \frac{hc}{\Delta E} = \frac{1240}{1.89} \text{ nm}

\boxed{\lambda \approx 656 \text{ nm}}

This is the famous H-alpha line — a red line in the Balmer series of hydrogen, visible to the naked eye.

Why This Works

Bohr’s model quantises angular momentum: $L = n\hbar$ . This restricts the electron to specific orbits with definite energies. The $1/n^2$ dependence arises from the balance between Coulomb attraction ( $\propto 1/r$ ) and centripetal acceleration, combined with the quantisation condition.

The ground state energy ( $-13.6$ eV) is one of the most important numbers in physics. It tells us how much energy is needed to ionise a hydrogen atom from its ground state — the ionisation energy.

For hydrogen-like ions (He $^+$ , Li $^{2+}$ , etc.), multiply by $Z^2$ : the energy of He $^+$ ground state is $-13.6 \times 4 = -54.4$ eV.

Alternative Method

Using the Rydberg formula directly:

\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

For $n_1 = 2$ , $n_2 = 3$ , $R_H = 1.097 \times 10^7$ m $^{-1}$ :

\frac{1}{\lambda} = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{5}{36} = 1.524 \times 10^6

$\lambda = 656 \text{ nm}$ — same answer.

For JEE, remember the series: Lyman ( $n_1 = 1$ , UV), Balmer ( $n_1 = 2$ , visible), Paschen ( $n_1 = 3$ , IR). The longest wavelength in each series comes from the smallest jump ( $n_2 = n_1 + 1$ ). The shortest wavelength (series limit) comes from $n_2 = \infty$ .

Common Mistake

When calculating $\Delta E$ , students sometimes subtract in the wrong order and get a negative wavelength. For emission, $\Delta E = E_{higher} - E_{lower}$ gives a positive value. The photon energy is always positive. Also, do not forget that $E_n$ values are negative — subtracting two negative numbers needs care with signs.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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