Physics, 22.06.2019 12:30 madlenserlipepu1o

Hydrogen atoms are excited by a laser to the =4n=4 state and then allowed to emit. what is the maximum number of distinct emission spectral lines (lines of different wavelengths) that can be observed from this system? 8 6 2 7 4 5 1 3 calculate the wavelength of the 4⟶14⟶1 transition. =λ=

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Answer from: bskyeb14579

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

Explanation:

The formula fro number of spectral lines are given by:

$l=\frac{(n_2-n_1)(n_2-n_1+1)}{2}$

n_1 =initial energy level

n_2 = final energy level.

Hydrogen atoms are excited by a laser to the n = 4 state .So the number of spectral lines observed will be:

$l=\frac{(4-1)(4-1+1)}{2}=6$

The wavelength of the 4 to 2 transition will be given by Rydberg expression:

$\frac{1}{\lambda }=R[\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}]$

$\frac{1}{\lambda }=1.097\times 10^{10}\times [\frac{1}{4^2}-\frac{1}{2^2}]$

$\lambda =486.17 nm$ ( $1nm =10^{-9} m$ )

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

Answer from: usjajdjajfj2ot929

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

Explanation:

The formula fro number of spectral lines are given by:

$l=\frac{(n_2-n_1)(n_2-n_1+1)}{2}$

n_1 =initial energy level

n_2 = final energy level.

Hydrogen atoms are excited by a laser to the n = 4 state .So the number of spectral lines observed will be:

$l=\frac{(4-1)(4-1+1)}{2}=6$

The wavelength of the 4 to 2 transition will be given by Rydberg expression:

$\frac{1}{\lambda }=R[\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}]$

$\frac{1}{\lambda }=1.097\times 10^{10}\times [\frac{1}{4^2}-\frac{1}{2^2}]$

$\lambda =486.17 nm$ ( $1nm =10^{-9} m$ )

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

Answer from: july00

The total number of spectral line for hydrogen atom at n=4 is $\boxed6$ .

The value of wavelength of a spectral line of transition from n=4 to n=2 is $\boxed{{\text{486}}{\text{.2 nm}}}$ .

Further explanation:

The total number of spectral line from a given initial energy level is,

${\text{Total number of spectral line}}=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}$ …… (1)

Here, n is the initial energy level of transition.

In the given question, the initial energy level of hydrogen atom is at 4, so substitute 4 for n in equation (1).

$\begin{aligned}{\text{Total number of spectral line}}&=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}\\&=\frac{{4\left({4-1}\right)}}{2}\\&=6\\\end{aligned}$

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, 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) 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)

Finding the wavelength of spectral lines in transition from n=4 to n=2

The initial energy level n=4 to final energy level n=2. Substitute the values of

$\begin{aligned}{\lambda _3}&=\frac{1}{{\left({1.097\times{{10}^7}{\text{ }}{{\text{m}}^{-1}}}\right)\left({\frac{1}{{{{\left({\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{4}}\right)}^2}}}}\right)}}\\&=4.862\times{10^{-7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{-9}}{\text{ m}}}}}\right)\\&=486.2{\text{ nm}}\\\end{aligned}$

Learn more:

1. Ranking of elements according to their first ionization

2. Chemical equation representing the first ionization energy for

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords:transition, hydrogen atom, energy difference, transition from n=4 to n=2, spectral lines, wavelength of spectral lines, and 486.2 nm.

Answer from: zoeycrew

The total number of spectral line for hydrogen atom at n=4 is $\boxed6$ .

The value of wavelength of a spectral line of transition from n=4 to n=2 is $\boxed{{\text{486}}{\text{.2 nm}}}$ .

Further explanation:

The total number of spectral line from a given initial energy level is,

${\text{Total number of spectral line}}=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}$ …… (1)

Here, n is the initial energy level of transition.

In the given question, the initial energy level of hydrogen atom is at 4, so substitute 4 for n in equation (1).

$\begin{aligned}{\text{Total number of spectral line}}&=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}\\&=\frac{{4\left({4-1}\right)}}{2}\\&=6\\\end{aligned}$

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)

Therefore, after rearrangement of equation (1) can be calculated as,

Finding the wavelength of spectral lines in transition from n=4 to n=2

The initial energy level n=4 to final energy level n=2. Substitute the values of

Learn more:

1. Ranking of elements according to their first ionization

2. Chemical equation representing the first ionization energy for

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords:transition, hydrogen atom, energy difference, transition from n=4 to n=2, spectral lines, wavelength of spectral lines, and 486.2 nm.

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Hydrogen atoms are excited by a laser to the =4n=4 state and then allowed to emit. what is the maxim...

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