Part A The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 1.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission? Express your answer numerically in joules.

Part B During an emission, electrons move from a higher energy orbital to a lower energy orbital. Which of the following are valid transitions that produce lines in the emission spectrum of Zn?

Check all that apply.

1-[Ar]4s13d106s1→[Ar]4s23d10

2-[Ar]4s23d10→[Ar]4s23d104p2

3-[Ar]4s23d10→[Ar]3d10

4-[Ar]4s23d10→[Ar]4s13d11

5-[Ar]3d10→[Ar]4s23d10

6-[Ar]4s23d10→[Ar]4s13d106s1

Respuesta :

Answer:

(a)

[tex]E=9.2887\times 10^{-9}}\ J[/tex]

(b)

1- [tex][Ar]4s^13d^{10}6s^1\rightarrow [Ar]4s^23d^{10}[/tex]

Explanation:

(a)

[tex]E=n\times \frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength

Given, [tex]\lambda=214\ nm=214\times 10^{-9}\ nm[/tex]

n is the number of atoms , [tex]1.00\times 10^{10}[/tex]

[tex]E=1.00\times 10^{10}\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{214\times 10^{-9}}\ J[/tex]

[tex]E=9.2887\times 10^{-9}}\ J[/tex]

(b)

Transition from higher level to lower level is : -

1- [tex][Ar]4s^13d^{10}6s^1\rightarrow [Ar]4s^23d^{10}[/tex]

Electron from 6s which has higher energy falls to 4s which has lower energy.

A line in the emission spectrum results from the movement of an electron from excited state back to its ground state.

We know the the energy of a photon is obtained from;

E = hc/λ

Where;

E = energy of the photon = ?

h = Plank's constant = 6.6 × 10^-34 Js

c = Speed of light = 3 × 10^8 m/s

λ = wavelength = 214 × 10^-9 m

Substituting values;

E = 6.6 × 10^-34 Js × 3 × 10^8 m/s/214 × 10^-9 m

E = 9.3  × 10^-19 J

Since there are 1.00×10^10 atoms of zinc emitting light,

E = 9.3  × 10^-19 J ×  1.00×10^10  = 9.3 × 10^-9 J

A line in the emission spectrum is produced when an electron returns to ground state from a higher energy excited state. This is only possible for the transition;  [Ar]4s23d10→[Ar]4s13d106s1

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