Answers from problem # 17 & 18

Given;
The energy of radiation, E=4.071E19J.
Provided; speed of electromagnetic radiation,c=3.0E8ms-¹.
Planck's constant,h=6.63E-34Js-¹.
Required;
(i) Wavelength of the green light,λ and
(ii) Higher energy level E2, from which the electron jumps to the ground energy level E1, i.e. n2.
Solution;
(i) From the planck's expression of energy.
E=hc
      λ
λ=hc
     E
⇒λ=6.63E-34Js×3.0E8ms-¹
                    4.071E¹19J
λ=4.888E7m.
.: Wavelength of the green light,λ=4.888E7m.

(ii) Recall the Rhydberg expression.
1 = RH (   1  _   1   )
λ              n₂²    n₁²
Where; RH is a constant= 1.097E7m-¹ ,n₂=n and n₁=2.
All other symbols carry their usual notations.
  2²n² = RH•λ
2² - n²
n²= 1.341E14 (4 - n²)
n²=5.362E14 - 1.341E14n²
1.341E14n²=5.362E14
⇒n=3.999≈4.
.: The higher energy level E2, from which the electron jumps to the ground energy level E1,n=4.

18.
Given;
The Peak heights of lead relative to their mass/charge ratios are 1.5↝204, 23.6↝206, 22.6↝207 and 52.3↝208.
Required;
The average atomic mass of lead, A.A.M.
Solution;
Recall;
A.A.M=Sum of the products of the peak values of lead and their relative abundances divided by the sum of these peak values.
⇒A.A.M=[(1.5×204)+(23.6×206)+(22.6×207)+(52.3×208)]÷[1.5+23.6+22.6+52.3]
A.A.M=207.242.
.:The average atomic mass of lead =207.242.
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