Answer
Verified
445.2k+ views
Hint: The hydrogen spectrum series is classified into five types. These are Lyman series, Balmer series, Paschen series, Bracket series and Pfund series.
Complete step by step answer:
We know that for Balmer series,
\[\dfrac{1}{{{\lambda }}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{\text{n}}_{\text{2}}^{\text{2}}}}} \right]\] …..(1)
For the alpha line, n = 3. Using this value in the above expression (1) we get,
\[\dfrac{1}{{{{{\lambda }}_a}}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{3^2}}}} \right] = \dfrac{1}{{\text{X}}}\]
\[
\Rightarrow \dfrac{1}{{{{{\lambda }}_a}}} = \,{{R \times }}\,\dfrac{{\text{5}}}{{{\text{36}}}} = \,\dfrac{1}{{\text{X}}} \\
\Rightarrow {\text{R}} = \,\dfrac{{{\text{36}}}}{{\text{5}}}{{ \times }}\dfrac{{\text{1}}}{{\text{X}}} \\
\] ……(2)
For the beta line, n = 4. Using this value in the above expression (1) we get,
\[ \Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right] = \,\dfrac{3}{{16}}{\text{R}}\,\]
\[ \Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{X}}\dfrac{{80}}{{108}}{{\text{A}}^{\text{0}}}\,\]……(3)
So, the correct answer is Option C.
Additional Information:
The general formula for the hydrogen emission spectrum is given by Johannes Rydbergis :
\[\mathop \nu \limits^ - = \,109677\,\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)\]
Here,
\[{n_1}\]= 1,2,3,4 …
\[{n_2}\] = n1 +1
\[\mathop \nu \limits^ - \]= wave number of the electromagnetic radiation.
The series can be classified according to the transition in the following way:
Transition from the first shell, i.e. (n=1) to any other shell gives rise to the Lyman series
Transition from the second shell i.e. (n=2) to any other shell gives rise to the Balmer series
Transition from the third shell i.e. (n=3) to any other shell gives rise to the Paschen series
Transition from the fourth i.e. (n=4) shell to any other shell gives rise to the Bracket series
Transition from the fifth shell i.e. (n=5) to any other shell gives rise to the Pfund series
Note:
The value of Rydberg constant for hydrogen is \[109,677\,{\text{c}}{{\text{m}}^{ - 1}}\]. The Rydberg constant is used for the calculation of the wavelengths in the hydrogen spectrum.
Complete step by step answer:
We know that for Balmer series,
\[\dfrac{1}{{{\lambda }}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{\text{n}}_{\text{2}}^{\text{2}}}}} \right]\] …..(1)
For the alpha line, n = 3. Using this value in the above expression (1) we get,
\[\dfrac{1}{{{{{\lambda }}_a}}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{3^2}}}} \right] = \dfrac{1}{{\text{X}}}\]
\[
\Rightarrow \dfrac{1}{{{{{\lambda }}_a}}} = \,{{R \times }}\,\dfrac{{\text{5}}}{{{\text{36}}}} = \,\dfrac{1}{{\text{X}}} \\
\Rightarrow {\text{R}} = \,\dfrac{{{\text{36}}}}{{\text{5}}}{{ \times }}\dfrac{{\text{1}}}{{\text{X}}} \\
\] ……(2)
For the beta line, n = 4. Using this value in the above expression (1) we get,
\[ \Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right] = \,\dfrac{3}{{16}}{\text{R}}\,\]
\[ \Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{X}}\dfrac{{80}}{{108}}{{\text{A}}^{\text{0}}}\,\]……(3)
So, the correct answer is Option C.
Additional Information:
The general formula for the hydrogen emission spectrum is given by Johannes Rydbergis :
\[\mathop \nu \limits^ - = \,109677\,\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)\]
Here,
\[{n_1}\]= 1,2,3,4 …
\[{n_2}\] = n1 +1
\[\mathop \nu \limits^ - \]= wave number of the electromagnetic radiation.
The series can be classified according to the transition in the following way:
Transition from the first shell, i.e. (n=1) to any other shell gives rise to the Lyman series
Transition from the second shell i.e. (n=2) to any other shell gives rise to the Balmer series
Transition from the third shell i.e. (n=3) to any other shell gives rise to the Paschen series
Transition from the fourth i.e. (n=4) shell to any other shell gives rise to the Bracket series
Transition from the fifth shell i.e. (n=5) to any other shell gives rise to the Pfund series
Note:
The value of Rydberg constant for hydrogen is \[109,677\,{\text{c}}{{\text{m}}^{ - 1}}\]. The Rydberg constant is used for the calculation of the wavelengths in the hydrogen spectrum.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers