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Question: What will be the longest wavelength line in the Balmer series of spectrum of \(H\) atom? \(\left( ...

What will be the longest wavelength line in the Balmer series of spectrum of HH atom?
(A)546nm\left( A \right)\,546nm
(B)656nm\left( B \right)656nm
(C)566nm\left( C \right)566nm
(D)556nm\left( D \right)\,556nm

Explanation

Solution

: Since wavelength and energy are inversely related ,the longest wavelength would be produced by the lowest amount of energy. All Balmer series lines have n1=2{n_1} = 2 ,hence the lowest amount is produced by n=3n = 3. By applying the wavelength formula for the Balmer series we can find the wavelength.

Complete step by step solution:
The Balmer series in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885.
We have to find the longest wavelength in the Balmer series as we all know that wavelength is inversely proportional to the energy .Hence in the Balmer series lowest amount is energy is produced by the principal quantum number which is equals to 3 i.e. n=3n = 3.
In the Balmer series n1=2{n_1} = 2and the lowest energy level is n2=3{n_2} = 3.
Now, by applying single electron species wavelength formula ,we get
1λ=RZ2(1n121n22)(1)\dfrac{1}{\lambda } = \,\,R{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) \cdot \cdot \cdot \cdot \left( 1 \right)
Where,
R = RydbergConstant{\text{R = Rydberg}}\,\,{\text{Constant}}
R=1.097×107m1\Rightarrow R = 1.097 \times {10^7}\,{m^{ - 1}}
Converting the above constant from m1tonm1{m^{ - 1}}\,to\,\,n{m^{ - 1}} by multiplying 109{10^{ - 9}},we get
R=1.097×107×109nm1R = 1.097 \times {10^7} \times {10^{ - 9}}\,n{m^{ - 1}}
R=1.097×102nm1\Rightarrow R = 1.097 \times {10^{ - 2}}\,\,n{m^{ - 1}}
Now For Hydrogen atoms, Z=atomicnumber=1Z = atomic\,\,number = 1
For the Balmer series, n1=2{n_1} = 2
And for the longest wavelength, minimum energy transition will occur at n2=3{n_2} = 3.
Now putting these values in equation (1)\left( 1 \right),we get
1λ=1.097×102nm1×12(122132)\dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}\,n{m^{ - 1}} \times {1^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)
1λ=1.097×102(1419)nm1\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)\,n{m^{ - 1}}
1λ=1.097×102(9436)nm1\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}\left( {\dfrac{{9 - 4}}{{36}}} \right)\,n{m^{ - 1}}
1λ=1.097×102(536)nm1\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}\left( {\dfrac{5}{{36}}} \right)\,n{m^{ - 1}}
1λ=5.48536×102nm1\Rightarrow \dfrac{1}{\lambda } = \dfrac{{5.485}}{{36}} \times {10^{ - 2}}\,n{m^{ - 1}}
1λ=0.152×102nm1\Rightarrow \dfrac{1}{\lambda } = 0.152 \times {10^{ - 2}}\,n{m^{ - 1}}
λ=6.56×102nm\Rightarrow \lambda = 6.56 \times {10^2}\,nm
λ=656nm\therefore\lambda = 656 nm
Therefore, the correct answer option is (B)\left( B \right).

Note: The value of Rydberg Constant is in m1{m^{ - 1}}but in our problem we have to find wavelength in nmnm,so first we have to convert the Rydberg Constant into nm1n{m^{ - 1}}by multiplying the constant value with 109{10^{ - 9}}.
(1m = 10 - 9nm)\left( {{\text{1}}\,{\text{m = 1}}{{\text{0}}^{{\text{ - 9}}}}\,{\text{nm}}} \right)