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Q.37

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Found in: Page 1296

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# a. What downward transitions are possible for a sodium atom in the $6s$ state? b. What are the wavelengths of the photons emitted in each of these transitions?

(a) $6s\to 5p,6s\to 4p$ and $6s\to 3p$

(b) $∆E=0.17eV$

See the step by step solution

## Diagram :

The allowed transitions are determined form the following usual selection rules

$∆L=±1\phantom{\rule{0ex}{0ex}}∆S=0\phantom{\rule{0ex}{0ex}}∆J=0,±1$

If the above selection rules are not satisfied, no other transitions are allowed between the energy levels.

The following diagram shows $\left[Ne\right]3s$ground state of sodium atom and some are in excited state.

## Part (a) Step 1:

The selection rule for a transition is

$∆l=±1$

Now for $s$ state $l=0$. The transition is possible if the lower state has $l$ value as $1$ . But for $6s$, there are possible lower states having $l=1$ are $5p,4p$ and $3p$.

The possible transitions are,

$6s\to 5p,6s\to 4p$ and $6s\to 3p$

## Part (b) Step 2:

Wavelength of photon emitted is

$\lambda =\frac{hc}{∆E}$

Rearranging the equation (1) for $∆E$.

$∆E=\frac{hc}{\lambda }$ ......(1)

Here, $h$ is Plank's constant $\left(4.14×{10}^{-15}eV.s\right)$ and $c$ is speed of light in vacuum $\left(3×{10}^{8}m/s\right)$.

The energy level difference in the transition, $6s\to 5p$ is,

$∆E={E}_{6s}-{E}_{5p}$

Substitute $4.51eV$ and ${E}_{6s}$ and $4.34eV$ for ${E}_{5p}$ in the above equation,

$∆E=4.51eV-4.34eV\phantom{\rule{0ex}{0ex}}=0.17eV$

## Substituting values in above equation:

Substitute $4.14×{10}^{-15}eV.s$ for $h,3×{10}^{8}m/s$ for $c$ and $0.17eV$ for $\lambda$ in the above equation.

localid="1649071820977" ${\lambda }_{6s\to 5p}=\frac{\left(4.14×{10}^{-15}eVs\right)\left(3×{10}^{8}m/s\right)}{\left(0.17eV\right)}\phantom{\rule{0ex}{0ex}}=\left(7300×{10}^{-9}m\right)\left(\frac{{10}^{9}nm}{1.0m}\right)\phantom{\rule{0ex}{0ex}}=7300nm$

Therefore the wavelength is $7300nm$

## Energy level difference 2:

The energy level difference in the transition $6s\to 4p$ is,

localid="1648557023475" $∆E={E}_{6s}-{E}_{4p}$

Substitute $4.51eV$ for ${E}_{6s}$ and $3.75eV$ for ${E}_{4p}$ in the above equation.

$∆E={E}_{6s}-{E}_{4p}\phantom{\rule{0ex}{0ex}}=4.51eV-3.75eV\phantom{\rule{0ex}{0ex}}=0.76eV$

Substitute localid="1648557182261" $4.14×{10}^{-15}eV.s$ for $h,3×{10}^{8}m/s$ for $c$ and $0.76eV$ for $\lambda$ in the above equation.

localid="1648557427578" ${\lambda }_{6s\to 5p}=\frac{\left(4.14×{10}^{-15}eVs\right)\left(3×{10}^{8}m/s\right)}{\left(0.7eV\right)}\phantom{\rule{0ex}{0ex}}=\left(1634×{10}^{-9}m\right)\left({10}^{9}nm/m\right)\phantom{\rule{0ex}{0ex}}=1634nm$

Therefore, the wavelength is $1634nm$.

## Energy level difference 3:

The energy level difference in the transition $6s\to 3p$ is,

$∆E={E}_{6s}-{E}_{4p}$

Substitute $4.51eV$ for ${E}_{5s}$ and $2.104eV$ for ${E}_{3p}$ in the above equation.

localid="1648557169920" $∆E=4.51eV-2.104eV\phantom{\rule{0ex}{0ex}}=2.406eV$

Substitute $4.14×{10}^{-15}eV.s$ for $h,3×{10}^{8}m/s$ for $c$ and $2.406eV$ for $\lambda$ in the above equation.

${\lambda }_{6s\to 5p}=\frac{\left(4.14×{10}^{-15}eVs\right)\left(3×{10}^{8}m/s\right)}{\left(2.406eV\right)}\phantom{\rule{0ex}{0ex}}=\left(516.2×{10}^{-9}m\right)\left({10}^{9}nm/m\right)\phantom{\rule{0ex}{0ex}}=516.2nm$

Therefore, the wavelength is $516.2nm$.

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