Question

Find out the wavelength of the next line in the series having lines of spectrum of H-atom of wavelengths \[656.46\] , $486.27$ , $434.17$ and $410.29{\text{nm}}$ .
A. $397{\text{nm}}$
B. ${\text{122nm}}$
C. ${\text{1282nm}}$
D. $302{\text{nm}}$

Answer 1

Hint: The reciprocal of wavelength, i.e., the wave number of a spectral line of a hydrogen atom can be calculated by using Rydberg's equation. 

According to this equation: $\dfrac{{\text{1}}}{{{\lambda }}}{\text{ = R}}\left( {\dfrac{{\text{1}}}{{{{\text{n}}_{\text{1}}}^{\text{2}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}}} \right)$
Here, R is a constant called the Rydberg’s constant, ${{\lambda }}$ is the wavelength and ${{\text{n}}_{\text{1}}}$ and ${{\text{n}}_2}$ are integers.
For a given spectral series, ${{\text{n}}_{\text{1}}}$ remains constant and ${{\text{n}}_2}$ varies for each line in the same series. For the Balmer series, the value of ${{\text{n}}_{\text{1}}}$ is equal to 2.

Complete answer:
The value of ${{\text{n}}_{\text{1}}}$ in a spectral series represents the lower energy level for that series and the value of ${{\text{n}}_2}$ represents the higher energy level for that series.
According to the question, the wavelengths of the lines of the given series lie within the visible region of the electromagnetic spectrum. And we know that the transitions corresponding to the Balmer series occurs in the visible region which has a wavelength range of 380 to 700 nm. Hence, the given series is the Balmer series.
So, the value of ${{\text{n}}_{\text{1}}}$ is 2. Let us find the value of ${{\text{n}}_2}$ for the last line of wavelength $410.29{\text{nm}}$ or $410.29 \times {10^{ - 7}}{\text{cm}}$ .
Also, the value of R is ${\text{109678c}}{{\text{m}}^{ - 1}}$ .
Using the Rydberg’s equation for Balmer series, we get:
\[
  \dfrac{{\text{1}}}{{410.29 \times {{10}^{ - 7}}}}{\text{ = 109678}}\left( {\dfrac{{\text{1}}}{{{2^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}}} \right) \\
   \Rightarrow 24300 = 27419.5 - \dfrac{{{\text{109678}}}}{{{{\text{n}}^{\text{2}}}}} \\
   \Rightarrow \dfrac{{{\text{109678}}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}} = 3119.5 \\
   \Rightarrow {{\text{n}}_{\text{2}}}^{\text{2}} = \dfrac{{{\text{109678}}}}{{3119.5}} \\
   \Rightarrow {{\text{n}}_{\text{2}}}^{\text{2}} = 35.16 \\
   \Rightarrow {{\text{n}}_{\text{2}}} = 6 \\
 \]
Hence, the next line will be from ${{\text{n}}_2} = 7$ to ${{\text{n}}_1} = 2$ .
Hence, the wavelength of this line is:
$
  \dfrac{{\text{1}}}{{{\lambda }}}{\text{ = 109678}}\left( {\dfrac{{\text{1}}}{{{2^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{7^{\text{2}}}}}} \right) \\
   \Rightarrow \dfrac{{\text{1}}}{{{\lambda }}}{\text{ = }}27419.5 - 2238.33 \\
   \Rightarrow \dfrac{{\text{1}}}{{{\lambda }}}{\text{ = 25181}}{\text{.17}} \\
   \Rightarrow {{\lambda }} = \dfrac{{\text{1}}}{{{\text{25181}}{\text{.17}}}} \\
   \Rightarrow {{\lambda }} = 3.97 \times {10^{ - 5}}{\text{cm}} \\
   \Rightarrow {{\lambda }} = 397{\text{nm}} \\
 $

So, the correct answer is A.

Note: 

> For the Lyman series, the lines occur in the ultraviolet region of the electromagnetic spectrum which has wavelengths 100 to 400 nm. The lines in these series appear when electrons drop from higher energy levels (${{\text{n}}_2} = 2,3,4$ etc.) to the first energy level (${{\text{n}}_1} = 1$ ). 

> For the Lyman series, the Rydberg’s equation will be:

$\dfrac{{\text{1}}}{{{\lambda }}}{\text{ = R}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}}} \right)$