Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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

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

(b) $∆ E = 0.17 e V$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Diagram :

The allowed transitions are determined form the following usual selection rules

$∆ L = \pm 1 ∆ S = 0 ∆ J = 0, \pm 1$

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

The following diagram shows $[N e] 3 s$ ground state of sodium atom and some are in excited state.

Part (a) Step 1:

The selection rule for a transition is

$∆ l = \pm 1$

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

The possible transitions are,

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

Part (b) Step 2:

Wavelength of photon emitted is

$λ = \frac{h c}{∆ E}$

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

$∆ E = \frac{h c}{λ}$ ......(1)

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

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

$∆ E = E_{6 s} - E_{5 p}$

Substitute $4.51 e V$ and $E_{6 s}$ and $4.34 e V$ for $E_{5 p}$ in the above equation,

$∆ E = 4.51 e V - 4.34 e V = 0.17 e V$

Substituting values in above equation:

Substitute $4.14 \times 10^{- 15} e V . s$ for $h, 3 \times 10^{8} m / s$ for $c$ and $0.17 e V$ for $λ$ in the above equation.

localid="1649071820977" $λ_{6 s \to 5 p} = \frac{(4.14 \times 10^{- 15} e V s) (3 \times 10^{8} m / s)}{(0.17 e V)} = (7300 \times 10^{- 9} m) (\frac{10^{9} n m}{1.0 m}) = 7300 n m$

Therefore the wavelength is $7300 n m$

Energy level difference 2:

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

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

Substitute $4.51 e V$ for $E_{6 s}$ and $3.75 e V$ for $E_{4 p}$ in the above equation.

$∆ E = E_{6 s} - E_{4 p} = 4.51 e V - 3.75 e V = 0.76 e V$

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

localid="1648557427578" $λ_{6 s \to 5 p} = \frac{(4.14 \times 10^{- 15} e V s) (3 \times 10^{8} m / s)}{(0.7 e V)} = (1634 \times 10^{- 9} m) (10^{9} n m / m) = 1634 n m$

Therefore, the wavelength is $1634 n m$ .

Energy level difference 3:

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

$∆ E = E_{6 s} - E_{4 p}$

Substitute $4.51 e V$ for $E_{5 s}$ and $2.104 e V$ for $E_{3 p}$ in the above equation.

localid="1648557169920" $∆ E = 4.51 e V - 2.104 e V = 2.406 e V$

Substitute $4.14 \times 10^{- 15} e V . s$ for $h, 3 \times 10^{8} m / s$ for $c$ and $2.406 e V$ for $λ$ in the above equation.

$λ_{6 s \to 5 p} = \frac{(4.14 \times 10^{- 15} e V s) (3 \times 10^{8} m / s)}{(2.406 e V)} = (516.2 \times 10^{- 9} m) (10^{9} n m / m) = 516.2 n m$

Therefore, the wavelength is $516.2 n m$ .

Find Study Materials for

Create Study Materials

Select your language

Physics For Scientists & Engineers

Answers without the blur.

Short Answer

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

Step by Step Solution

Diagram :

Part (a) Step 1:

Part (b) Step 2:

Substituting values in above equation:

Energy level difference 2:

Energy level difference 3:

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Physics For Scientists & Engineers

Answers without the blur.

Short Answer

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?

Step by Step Solution

Diagram :

Part (a) Step 1:

Part (b) Step 2:

Substituting values in above equation:

Energy level difference 2:

Energy level difference 3:

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics

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