Chemistry, 17.09.2019 12:20 lizzyhearts

The spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. calculate the wavelengths associated with the spectral transitions of the hydrogen atom from the n=6,5,4 and 3 to the n=2 level.

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Answer from: dondre54

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

Answer from: mom9032

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:

1. Ranking of elements according to their first ionization energy.:

2. Chemical equation representing the first ionization energy for lithium.:

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.

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