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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