Question

A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident radiation should be:
A. $800\mathop A\limits^o $
B. $825\mathop A\limits^o $
C. $975\mathop A\limits^o $
D. $1025\mathop A\limits^o $

Answer 1

Hint: As we know that the Number of Spectral Lines defines an element's absorption spectrum that has dark lines in the same positions as its emission spectrum's bright lines. 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:
$\text{Number of spectral lines} = \dfrac{{n(n - 1)}}{2}$
The expression of the energy of the radiation according to the Rydberg equation that corresponds to the transition between two energy levels is given by:
$E = - 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$,
And, the energy which is necessary for the transition of an electron is given by:
$E = \dfrac{{hc}}{\lambda }$.
Here, $h$ is the Planck constant and $c$ is the speed of light.

Complete step by step solution:
In the question, we have given the number of spectral lines as $6$. 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 formula for finding the total number of possible spectral lines,
$\text{Number of spectral lines}= \dfrac{{n(n - 1)}}{2}$

Now, substitute the given number of spectral lines in the above formula, then we obtain:
$6 = \dfrac{{n(n - 1)}}{2} \\$
$\Rightarrow 12 = {n^2} - n \\$
Factorise the above equation with the help of middle term splitting, then we have:
${n^2} - 4n + 3n - 12 = 0 \\$
$\Rightarrow n(n - 4) + 3(n - 4) = 0 \\$
$\Rightarrow (n + 3)(n - 4) = 0 \\$
$\Rightarrow n = - 3,4 \\$
As $n$ should be positive, we consider the value of $n = 4$. So, the electron needs to reach the third excited state in order to generate $6$ emission lines.

In Bohr's hydrogen atom model, as an electron moves from a higher energy state to a lower energy state, it emits a photon with energy equal to the difference between the final and beginning energy states,
$\Delta E = {E_f} - {E_i}$
The hydrogen atom needs the energy to be stimulated to $4$ state from its ground state, then we have:
$\Delta E = {E_4} - {E_1}$

According to the Rydberg equation, the energy gain needed for a transition to the first excited state is as follows:
$\Delta E = - 13.6eV\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{1^2}}}} \right) \\$
$\Rightarrow \Delta E = - 13.6eV\left( {\dfrac{1}{{16}} - 1} \right) \\$
$\Rightarrow \Delta E = - 13.6eV\left( {\dfrac{{ - 15}}{{16}}} \right) \\$
$\Rightarrow \Delta E = 12.75\,eV $

To identify the wavelength, we can disregard the negative sign, which suggests that energy was released during the transition. Then, we can keep in mind that the energy and wavelength are related by:
$\Delta E = \dfrac{{hc}}{\lambda }$
From the above formula, we have:
$12.75eV = \dfrac{{hc}}{\lambda } $
As we know that the value of $hc$ in $eV$ comes out to be $1.23984193 \times {10^{ - 6}}eV$, so we consider the value of $hc$ as $12425eV\mathop A\limits^o $, then we obtain:
$\lambda = \dfrac{{12425eV\mathop A\limits^o }}{{12.75eV}} \\$
$\therefore \lambda = 975\mathop A\limits^o $

Thus, the correct option is C.

Note: An atom's electrons make transition from lower to higher energy levels as a result of energy absorption. Since the excited state of these excited electrons is unstable, they must emit energy in order to return to ground states. The frequencies of this light's emission make up the emission spectrum. On the other hand, when energy is absorbed by the electrons in the ground state to reach higher energy levels, an absorption spectrum is formed by the frequencies of light transmitted with dark bands.