Q.Using an expression for energy of electron, obtain the Bohr's formula for hydrogen spectral lines. OR State the law of radioactive decay. Hence derive the relation $N = N_0 e^{-\lambda t}$. Represent it graphically.

Bohr's energy expression for the H atom gives the Rydberg/spectral-line formula; separately, radioactive decay follows an exponential law derived from $dN/dt \propto -N$.

Option — Bohr's formula for hydrogen spectral lines

For the hydrogen atom, the total energy of the electron in the $n$-th orbit is

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

When an electron makes a transition from a higher orbit $n_2$ to a lower orbit $n_1$ ($n_2>n_1$), a photon is emitted with energy equal to the difference:

$h\nu = E_{n_2}-E_{n_1} = -\frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) = \frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Since $\nu = c/\lambda$,

$\frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2h^3c}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where $R = \dfrac{me^4}{8\varepsilon_0^2h^3c}$ is the Rydberg constant. This is Bohr's formula for the hydrogen spectral lines: for fixed $n_1$ and varying $n_2 = n_1+1, n_1+2,\dots$, it generates a whole series of spectral lines (Lyman for $n_1=1$, Balmer for $n_1=2$, etc.).

— OR (alternative) —

Law of radioactive decay: The number of undecayed radioactive nuclei decreases with time such that the rate of decay ($-dN/dt$) at any instant is directly proportional to the number of undecayed nuclei $N$ present at that instant:

$-\frac{dN}{dt} \propto N \quad\Rightarrow\quad -\frac{dN}{dt} = \lambda N$

where $\lambda$ is the decay constant (a characteristic of the radioactive nuclide).

Derivation of $N=N_0e^{-\lambda t}$:

$\frac{dN}{N} = -\lambda\,dt$

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