Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 7, Problem 7.132AP

 (a)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of energies lost in the two-step process is same as the energy lost in the single transition from n=3 to n=1 ”.

 (a)

Expert Solution
Check Mark

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

Given

The transition occurs directly from n=3 to n=1 .

The transition occurs first from energy level n=3 to n=2 and then from n=2 to n=1 .

The energy difference between two energy levels is given as,

ΔE=RHZ2(1nf21ni2) (1)

Where,

  • RH is the Rydberg constant (2.18×1018J) .
  • nf is the final energy level.
  • ni is the initial energy level.
  • ΔE is the energy difference.
  • Z is the atomic number.

The energy lost in one step process is given as,

Substitute the value of nf=1 , ni=3 and RH in the equation (1).

ΔE=RHZ2(1nf21ni2)=(2.18×1018J)(2)2(1(1)21(3)2)=7.75×10-18J

The negative sign indicates the loss of energy.

The energy lost in two step process is given as,

Substitute the value of nf=2 , ni=3 , Z and RH in the equation (1).

ΔE1=RHZ2(1nf21ni2)=(2.18×1018J)(2)2(1(2)21(3)2)=1.21×10-18J

Substitute the value of nf=1 , ni=2 , Z and RH in the equation (1).

ΔE2=RHZ2(1nf21ni2)=(2.18×1018J)(2)2(1(1)21(2)2)=6.54×10-18J

The total energy lost in two step process =ΔE1+ΔE2=((1.21×1018)(6.54×1018))J=7.75×1018J

Therefore, the value of energy lost is same whether the transition occurs in one step process or in two step process.

(b)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of wavelengths of the two photons emitted in the two-step process is equal to the wavelength of the single photon emitted in the transition from n=3 to n=1 ”.

(b)

Expert Solution
Check Mark

Answer to Problem 7.132AP

Solution

The given statement is false.

Explanation of Solution

Explanation

The relation between energy, wavelength and speed of light is given as,

ΔE=hcλ (2)

Where,

  • h is the Planck’s constant (6.626×1034Js) .
  • λ is the wavelength of the light.
  • c is the speed of the light (3×108m/s) .
  • ΔE is the energy difference.

The wavelength of photon emitted in one step transition is given as,

Substitute the value of h , c and ΔE in the equation (2),

ΔE=hcλ7.75×1018J=6.626×1034Js×3×108msλ7.75×1018J=19.878×1026Jmλλ=2.56×108m

The wavelength that corresponds to given transition is 2.56×108m .

The wavelength of photon emitted in two step transition is given as,

Substitute the value of h , c and ΔE in the equation (2),

ΔE=hcλ11.21×1018J=6.626×1034Js×3×108msλ11.21×1018J=19.878×1026Jmλ1λ1=16.43×108m

Substitute the value of h , c and ΔE in the equation (2),

ΔE=hcλ26.54×1018J=6.626×1034Js×3×108msλ26.54×1018J=19.878×1026Jmλ2λ2=3.04×108m

The total wavelength of light in two step process =λ1+λ2=((16.43×108)+(3.04×1018))J=19.47×108m

Therefore, the wavelengths that are emitted in two step process is not similar to wavelength of photons emitted in one step process.

(c)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of frequencies of the two photons emitted in the two-step process is equal to the frequency of the single photon emitted in the transition from n=3 to n=1 ”.

(c)

Expert Solution
Check Mark

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

The relation between frequency, wavelength and speed of light is given as,

ν=cλ (3)

Where,

  • ν is frequency of the light.
  • λ is the wavelength of the light.
  • c is the speed of the light (3×108m/s) .

The frequency of photon emitted in one step transition is given as,

Substitute the value of λ , c in the equation (2),

ν=cλ=3×108m/s2.56×108m=1.17×1016s1

The frequency that corresponds to given transition is 1.17×1016s1 .

The frequency of photon emitted in two step transition is given as,

Substitute the value of λ , c in the equation (2),

ν1=cλ=3×108m/s16.43×108m=0.18×1016s1

Substitute the value of λ , c in the equation (2),

ν2=cλ=3×108m/s3.04×108m=0.99×1016s1

The total frequency of photons in two step process =ν1+ν2=((0.18×1016)+(0.99×1016))J=1.17×1016s1

Therefore, the frequencies that are emitted in two step process are similar to wavelength of photon emitted in one step process.

(d)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The wavelength of photon emitted by the He+ ion in the n=3 to n=1 transition is shorter than the wavelength of the photon emitted by H atom in an n=3 to n=1 transition”.

(d)

Expert Solution
Check Mark

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

The wavelength of photon emitted by He+ ion during transition from n=3 to n=2 is calculated as 2.56×108m .

The energy of Hydrogen atom is calculated by substituting the value of nf=1 , ni=3 , Z=1 and RH in the equation (1).

ΔE=RHZ2(1nf21ni2)=(2.18×1018J)(1)2(1(1)21(3)2)=1.94×10-18J

Substitute this value of h , c and ΔE in the equation (2),

ΔE=hcλ1.94×1018J=6.626×1034Js×3×108msλ1.94×1018J=19.878×1026Jmλλ=10.24×108m

The wavelength of photon emitted by the He+ ion in the n=3 to n=1 transition is shorter than the wavelength of the photon emitted by H atom in an n=3 to n=1 transition.

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Chapter 7 Solutions

Chemistry

Ch. 7.8 - Prob. 11PECh. 7.9 - Prob. 12PECh. 7.9 - Prob. 13PECh. 7.10 - Prob. 14PECh. 7 - Prob. 7.1VPCh. 7 - Prob. 7.2VPCh. 7 - Prob. 7.3VPCh. 7 - Prob. 7.4VPCh. 7 - Prob. 7.5VPCh. 7 - Prob. 7.6VPCh. 7 - Prob. 7.7VPCh. 7 - Prob. 7.8VPCh. 7 - Prob. 7.9VPCh. 7 - Prob. 7.10VPCh. 7 - Prob. 7.11VPCh. 7 - Prob. 7.12VPCh. 7 - Prob. 7.13VPCh. 7 - Prob. 7.14VPCh. 7 - Prob. 7.15VPCh. 7 - Prob. 7.16VPCh. 7 - Prob. 7.17VPCh. 7 - Prob. 7.18VPCh. 7 - Prob. 7.19VPCh. 7 - Prob. 7.20VPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - Prob. 7.42QPCh. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Prob. 7.100QPCh. 7 - Prob. 7.101QPCh. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Prob. 7.112QPCh. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - Prob. 7.119QPCh. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Prob. 7.123QPCh. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Prob. 7.126QPCh. 7 - Prob. 7.127QPCh. 7 - Prob. 7.128QPCh. 7 - Prob. 7.129APCh. 7 - Prob. 7.130APCh. 7 - Prob. 7.131APCh. 7 - Prob. 7.132APCh. 7 - Prob. 7.133APCh. 7 - Prob. 7.134APCh. 7 - Prob. 7.135APCh. 7 - Prob. 7.136APCh. 7 - Prob. 7.137APCh. 7 - Prob. 7.138APCh. 7 - Prob. 7.139APCh. 7 - Prob. 7.140APCh. 7 - Prob. 7.141APCh. 7 - Prob. 7.142APCh. 7 - Prob. 7.143APCh. 7 - Prob. 7.144APCh. 7 - Prob. 7.145APCh. 7 - Prob. 7.146APCh. 7 - Prob. 7.147APCh. 7 - Prob. 7.148AP
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