Question

In H-atom, electrons transit from 6th orbit to 2nd orbit in multi-step, then total spectral lines (without Balmer) will be:
A. 6
B. 10
C. 4
D. 0

Answer 1

Hint: According to Bohr when an atom makes a transition from higher energy level to a lower level it emits a photon with energy equal to energy difference between the initial and final levels. From the nth state, the atom may go to (n-1) th state… 2nd state or 1st state.

Complete step by step solution:
First let us discuss the concept behind the formula which we will apply here.
There is energy difference between Bohr orbits and therefore the wavelength of emitted or absorbed photon is given by the formula: \[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)\]
Where \[{n_2}\]is the upper energy level and \[{n_1}\]is the lower energy level and R is the Rydberg constant\[1.097 \times {10^7}{m^{ - 1}}\]
\[{n_1}\]=1-Lyman Series: All the wavelengths in the Lyman series are in the ultraviolet band.
\[{n_1}\]=2-Balmer Series: All the wavelengths in the Balmer series in the visible range. It consists of lines like H-alpha, H-beta, H-gamma and H-alpha line used in astronomy to detect the presence of hydrogen.
\[{n_1}\]=3-Paschen Series: The paschen lines all lie in the infrared band.
\[{n_1}\]=4-Brackett Series,
\[{n_1}\]=5-Pfund Series
Now let us move to the formula which we have to use in this problem.
The total number of spectral lines emitted on transition from one orbit to another are;
\[{n_2}\]where $n_1$ is the orbit to which transition taking place and $n_2$ be the orbit from which transition is taking place.
Now let’s apply the formula:
Given $n_1$ =2, $n_2$ =6
\[ = \dfrac{{(6 - 2)(6 - 2 + 1)}}{2}\]
\[ = 10\]
Now number of balmer line possible are 4 :$6—2,5—2,4—2,3—2 $
So there are 4 lines possible which we have to subtract from the total lines emitted as given in question.
So the total number of lines emitted without balmer are.

Hence our correct option is: A. 6

Note:
Remember the formula correctly and don’t confuse between \[{n_1}\] and \[{n_2}\] values. Read the question carefully as in this question because 10 might be the answer but if we have read the question correctly then we come to know that 6 is the correct answer.