The energy needed to remove an electron from n = 1 and n = 5 in the Li 2+ ion and the wavelength of the emitted photon in a transition from n = 5 to n = 1 are to be determined. Concept introduction: When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of electron. The transition energy ( Δ E ) is determined as follows: Δ E = − 2.18 × 10 − 18 J ( 1 n f 2 − 1 n i 2 ) Here, n f is the excited state orbital, and n i is the ground state orbital. The wavelength of a photon during a transition from one state to another can be expressed as follows: 1 λ = 2.18 × 10 − 18 J h c ( 1 n f 2 − 1 n i 2 ) ( because Δ E = h γ = hc/ λ ) Here, n f is the excited state, and n i is the ground state, h is Planck’s constant ( 6.63 × 10 − 34 J ⋅ s ) , c is the speed of light ( 3.0 × 10 8 m/s ) , and λ is the wavelength.

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Question

Chapter 6, Problem 118AP

Interpretation Introduction

Interpretation:

The energy needed to remove an electron from n=1 and n=5 in the Li2+ ion and the wavelength of the emitted photon in a transition from n=5 to n=1 are to be determined.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of electron. The transition energy (ΔE) is determined as follows:

ΔE=−2.18×10−18 J(1nf2−1ni2)

Here, nf is the excited state orbital, and ni is the ground state orbital.

The wavelength of a photon during a transition from one state to another can be expressed as follows:

1λ=2.18×10−18 Jhc(1nf2−1ni2)

( because ΔE = hγ = hc/λ)

Here, nf is the excited state, and ni is the ground state, h is Planck’s constant (6.63×10−34 J⋅s), c is the speed of light (3.0×108 m/s), and λ is the wavelength.

Expert Solution & Answer

Answer to Problem 118AP

Solution: The energies needed to remove an electron from n=1 and n=5 in the Li2+ ion are 1.96×10−17 J and 7.85×10−19 J. The wavelength of the emitted photon in a transition is 10.6 nm

Explanation of Solution

Given information:

ΔE=R∞hc(1n12−1n22)R∞=1.097×107 m−1

n1=5 and n2=1

ni=5 and nf=1

The difference in energy levels is calculated by the expression given below.

ΔE=−R∞hc(1n12−1n22) …… (1)

The Rydberg constant for hydrogen-like atoms R∞Z2= (1.097×107m−1)Z2.

For ni=1,nf=∞, and Z=3, the difference in energy is calculated by using equation (1) as follows:

ΔE=−(1.097×107m−1)(3)2(6.63×10−34J.s)(3.00×108m/s)(1∞2−112)=1.96×10−17 J

For ni=5,nf=∞, and Z=3, the difference in energy is calculated by using equation (1) as follows:

ΔE=−(1.097×107m−1)(3)2(6.63×10−34J.s)(3.00×108m/s)(1∞2−152)=7.85×10−19 J

The energies needed to remove an electron from n=1 and n=5 in the Li2+ ion are 1.96×10−17 J and 7.85×10−19 J.

Calculation of ΔE by using equation (1) as follows: ΔE=−(1.097×107m−1)(3)2(6.63×10−34J.s)(3.00×108m/s)(112−152)=−1.88×10−17 J

The wavelength of the emitted photon in the electronic transition from n=5 to n=1 is calculated by using the expression as follows:

λ = hcΔE

Substitute the values of h, c and ΔE.

λ = (6.63 × 10−34 J⋅s)(3.00 × 108 m/s)1.88 × 10−17 J=1.06×10−8m=10.6 nm

Conclusion

The energies needed to remove an electron from n=1 and n=5 in the Li2+

ion are 1.96×10−17 J

and 7.85×10−19 J respectively. The wavelength of the emitted photon in a transition is 10.6 nm.

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Answer 2