Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 7, Problem 7.64QP
Interpretation Introduction

Interpretation: The transitions from higher energy states to the n=2 level in He+ ever produce visible light is to be stated. The value of n2 at which visible light is produced is to be stated.

Concept introduction: When an atom or a molecule makes electronic transition from a high energy state to a lower energy state, the spectrum of different frequencies of electromagnetic radiations is obtained known as emission spectrum.

To determine: The transitions from higher energy states to the n=2 level in He+ ever produce visible light or not. The value of n2 at which visible light is produced.

Expert Solution & Answer
Check Mark

Answer to Problem 7.64QP

Solution

The transitions from higher energy states to the n=2 level in He+ will not produce visible light.

Explanation of Solution

Explanation

Given

The transition occurs from higher energy state to n=2 level.

The energy of the photons emitted by one electron atoms and ions is given by the equation,

E=(2.18×1018J)Z2(1n121n22) (1)

Where,

  • E is the energy of photons.
  • Z is an atomic number.
  • n1 is the lower energy state.
  • n2 is the higher energy state.

The values of n1 and n2 are considered to be 2 and respectively.

Substitute the values of Z , n1 and n2 in the above equation to calculate the energy of photons.

E=(2.18×1018J)(2)2(1(2)21()2)=(2.18×1018J)×4×(141)=8.72×1018×(0.250)=2.18×1018J

Hence, energy of the photons is 2.18×1018J .

The wavelength of the photons is calculated by using the equation,

λ=hcE (2)

Where,

  • λ is the wavelength.
  • h is the Planck’s constant (6.62×1034J.s) .
  • c is the speed of light (3×108m/s) .
  • E is the energy.

Substitute the values of E , h and c in the above equation to calculate the wavelength of photons.

λ=(6.62×1034J.s)(3×108m/s)2.18×1018J=1.986×1025J.m2.18×1018J=9.110×108m

The conversion of m to nm is done as,

1m=109nm

Therefore, the conversion of 9.110×108m to nm is done as,

9.110×108m=9.110×108×109nm=91.10nm

In the continuous spectrum of electromagnetic radiation, the wavelength of visible region ranges from 390nmto750nm . Here, the wavelength of photons is 91.10nm that will occur in the ultra-violet region.

Hence, transitions from higher energy states to the n=2 level in He+ will not produce visible light.

The transition in the visible light is produced when the value of wavelength must range from 390nmto750nm . The calculated value of wavelength is 91.10nm , hence to increase the value of wavelength the value of energy of electromagnetic radiation is to be decreases. As, it is known that,

E(1n121n22)

For lower value of E , the value of (1n121n22) must be low. As the increase in value of n2 , increases the value of (1n121n22) . Hence, the value of n2 must be low to have wavelength of visible region. Therefore, the values of n1 and n2 are considered to be 2 and 3 respectively. The energy of photon is calculated by using the equation (1).

Substitute the values of Z , n1 and n2 in equation (1) to calculate the energy of photons.

E=(2.18×1018J)(2)2(1(2)21(3)2)=(2.18×1018J)×4×(1419)=8.72×1018×(0.250.11)=8.72×1018×0.14J

Simplify the above equation,

E=1.220×1018J

The wavelength of photons emitted in the transition from n=3ton=2 is calculated by using the equation (2).

Substitute the values of E , h and c in the above equation to calculate the wavelength of photons.

λ=(6.62×1034J.s)(3×108m/s)1.22×1018J=1.986×1025J.m1.22×1018J=16.27×108m

The conversion of m to nm is done as,

1m=109nm

Therefore, the conversion of 16.27×108m to nm is done as,

16.27×108m=16.27×108×109nm=162.7nm

The wavelength of photons is still lesser than the wavelength required producing visible light. Hence, transitions from higher energy states to the n=2 level in He+ will never produce visible light.

Conclusion

Transitions from higher energy states to the n=2 level in He+ will never produce visible light as the calculated wavelength is lesser than the wavelength required producing visible light.

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Chapter 7 Solutions

Chemistry

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