Verified Answer :
Rydberg formula is given by:
\(\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )\)
where, \(R_{H}\) = Rydberg constant = \(1.0973731568508 \times 10^{7} per metre\)
\(\lambda\) = wavelength
\(n_{1}\) and \(n_{2}\) are the level of transitions.
Now, for \(n_{1}\)= 2 and \(n_{2}\)= 6
\(\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )\)
= \(1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )\)
= \(1.0973731568508 \times 10^{7} \times (0.25-0.0278 )\)
= \(1.0973731568508 \times 10^{7} \times 0.23\)
= \(0.2523958\times 10^{7}\)
\(\lambda = \frac{1}{0.2523958\times 10^{7}}\)
= \(3.9620\times 10^{-7} m\)
= \(396.20\times 10^{-9} m\)
= \(396.20 nm\)
Now, for \(n_{1}\)= 2 and \(n_{2}\)= 5
\(\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )\)
= \(1.0973731568508 \times 10^{7} \times (0.25-0.04 )\)
= \(1.0973731568508 \times 10^{7} \times (0.21 )\)
= \(0.230 \times 10^{7}\)
\(\lambda= \frac{1}{0.230 \times 10^{7}}\)
= \(4.3478 \times 10^{-7} m\)
= \(434.78\times 10^{-9} m\)
= \(434.78 nm\)
Now, for \(n_{1}\)= 2 and \(n_{2}\)= 4
\(\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )\)
= \(1.0973731568508 \times 10^{7} \times (0.25-0.0625 )\)
= \(1.0973731568508 \times 10^{7} \times (0.1875 )\)
= \(0.20575 \times 10^{7}\)
\(\lambda= \frac{1}{0.20575 \times 10^{7}}\)
= \(4.8602 \times 10^{-7} m\)
= \(486.02 \times 10^{-9} m\)
= \(486.02 nm\)
Now, for \(n_{1}\)= 2 and \(n_{2}\)= 3
\(\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )\)
= \(1.0973731568508 \times 10^{7} \times (0.25-0.12 )\)
= \(1.0973731568508 \times 10^{7} \times (0.13 )\)
= \(0.1426585\times 10^{7}\)
\(\lambda= \frac{1}{0.1426585\times 10^{7}}\)
= \(7.0097 \times 10^{-7} m\)
= \(700.97 \times 10^{-9} m\)
= \(700.97 nm\)
The wavelengths of spectral line observed in hydrogen atom are,
The value of wavelength of first spectral line from n=6 to n=2 is \(\boxed{{\text{410}}{\text{.2 nm}}}\).
The value of wavelength of second spectral line from n=5 to n=2 is \(\boxed{{\text{434}}{\text{.1 nm}}}\) .
The value of wavelength of third spectral line from n=4 to n=2 is \(\boxed{{\text{486}}{\text{.2 nm}}}\) .
The value of wavelength of fourth spectral line from n=3 to n=2 is \(\boxed{{\text{656}}{\text{.3 nm}}}\) .
Further explanation:
Concept:
According to the Rydberg equation, the wavelength of spectral line related with the transition values as follows:
\(\frac{1}{\lambda }=\left( {{{\text{R}}_{\text{H}}}}\right)\left({\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}}\right)\) …… (1)
Here, \(\lambda\) is the wavelength of spectral line, \({{\text{R}}_{\text{H}}}\) is the Rydberg constant that has the value \(1.097\times{10^7}{\text{ }}{{\text{m}}^{-1}}\) , \({{\text{n}}_{\text{i}}}\) is the initial energy level of transition, and \({{\text{n}}_{\text{f}}}\) is the final energy level of transition.
Therefore, after rearrangement of equation (1) \(\lambda\) can be calculated as,
\(\lambda=\frac{1}{{\left({1.097\times {{10}^7}{\text{ }}{{\text{m}}^{-1}}}\right)\left( {\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}}\right)}}\) …… (2)
Solution:
Finding the wavelength of spectral lines in each transition.
1. For the first transition, from initial energy level n=6 to final energy level n=2.
\(\begin{aligned}{\lambda _1}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{6}}\right)}^2}}}}\right)}}\\&= 4.102\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{410}}{\mathbf{.2 nm}}\\\end{aligned}\)
2. For the second transition, from initial energy level n=5 to final energy level n=2.
\(\begin{aligned}{\lambda _2}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{5}}\right)}^2}}}}\right)}}\\&= 4.341\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{434}}{\mathbf{.1 nm}}\\\end{aligned}\)
3. For the third transition, from initial energy level n=4 to final energy level n=2.
\(\begin{aligned}{\lambda _3}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{4}}\right)}^2}}}}\right)}}\\&= 4.862\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{486}}{\mathbf{.2 nm}}\\\end{aligned}\)
4. For the fourth transition, from initial energy level n=3 to final energy level n=2.
\(\begin{aligned}{\lambda _4}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{3}}\right)}^2}}}}\right)}}\\&= 6.563\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{656}}{\mathbf{.3 nm}}\\\end{aligned}\)
Learn more:
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2. Chemical equation representing the first ionization energy for lithium.: https://brainly.com/question/5880605
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Atomic structure
Keywords: transition, hydrogen atom, energy difference, transition from n=6 to n=2, transition from n=5 to n=2, transition from n=4 to n=2, transition from n=3 to n=2, spectral lines, wavelength of spectral lines.