Question

The ionisation potential of H-atom is $13.6V$. When it is excited from ground state by monochromatic radiations of $970.6\mathop A\limits^o $, the number of emission lines will be (according to Bohr's theory):
A. $10$
B. $8$
C. $6$
D. $4$

Answer 1

Hint:As we know that the Hydrogen Theory of Niels Bohr, according to Atom and Hydrogen-like Atoms, a hydrogen-like atom is made up of a tiny positively charged nucleus and an electron that revolves around the nucleus in a stable circular orbit. In order to solve the question, we should remember the principles of the hydrogen spectrum and the number of spectral lines in a given series represents all potential transitions from an energy level that is greater than that level to the energy level that corresponds to that series.

Formula used:
The following formula can be used to determine how many spectral lines or emission lines the electrons will produce when they fall from orbit $n$ to the ground state:
$\text{Number of emission lines} = \dfrac{{n(n - 1)}}{2}$
The wavelength for the hydrogen spectrum's corresponding series is given by: $\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$,
Here, $\lambda $ is the wavelength, $R$ is the Rydberg constant.

Complete step by step solution:
In the question, we have given that H-atom has an ionisation potential of $13.6V$. To calculate the orbit where the ground state electron begins to fall that will be produced when the electron transitions from the $n$ to the ground state, use the Rydberg equation, the energy gain needed for a transition to the first excited state. The wavelength for the hydrogen spectrum's corresponding series is given by:
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\,\,\,\,\,...(i)$

The hydrogen atom needs the energy to be stimulated to ${n_1} = 1$ state from its ground state and wavelength,
$\lambda = 970.6\mathop A\limits^o \\
\Rightarrow \lambda = 970.6 \times {10^{ - 10}}$
And, as we know, the Hydrogen Rydberg Constant value is $R \approx 1.097 \times {10^7}{m^1}$.

Now, substitute the above information in the formula $(i)$, then we have:
$\dfrac{1}{{970.6 \times {{10}^{ - 10}}}} = 1.097 \times {10^7}\left( {\dfrac{1}{{{{(1)}^2}}} - \dfrac{1}{{n_2^2}}} \right) \\$
$\Rightarrow \dfrac{1}{{970.6 \times {{10}^{ - 10}}}} = \left( {1.097 \times {{10}^7} - \dfrac{{1.097 \times {{10}^7}}}{{n_2^2}}} \right) \\$
$\Rightarrow {n_2} = \sqrt {16} = 4 \\$
Now, use the formula for finding the total number of possible emission lines,
$\text{Number of emission lines} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the obtained value of $n$ in the above formula, then we obtain:
$\text{Number of emission lines} = \dfrac{{4(4 - 1)}}{2} \\$
$\therefore \text{Number of emission lines} =\dfrac{{4 \times 3}}{2} = 6 \\$

Thus, the correct option is C.

Note: Atoms in a gaseous form emit light at discrete wavelengths with black intervals in between, rather than exhibiting a continuous spectrum and an excited electron might return both immediately and in stages. The energy released, however, won't change once it reaches its ground state.