Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom. ???
Answers
Answer:
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Explanation:
Second excited state (n=3)
Third excited state (n=4)
We know that,
E=hcλ13.6n2=hcλ13.6n2hc=1λ λ= n2hc13.6
Now,
λ2λ3= n22hc13.6×13.6n32hcλ2λ3=n22n23=(3)2(4)2λ2λ3=916
λ2 : λ3 = 9 : 16
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Concept:
The energy of the atom can be represented as,
E = hc/λ
where h is Planck's constant, c is the speed of the light and λ is the de-Broglie wavelength.
The energy can also be represented as E = 13.6/ n².
Given:
The electrons are in the second and third excited states.
Find:
The ratio of the de-Broglie wavelengths.
Solution:
As for the second excited state, n₁ = 3, and for the third excited state, n₂ = 4.
E = hc/λ and E = 13.6/ n².
So, de-Broglie wavelength and excited states can be related as,
hc/λ = 13.6/ n²
So, for n₁ = 3, λ = λ₁
hc/λ₁ = 13.6/ n₁² ...(i)
and for n₂ = 4, λ = λ₂
hc/λ₂ = 13.6/ n₂² ...(ii)
Taking the ratio of i and ii,
(hc/λ₁)/(hc/λ₂) = (13.6/ n₁² )/ (13.6/ n₂²)
Cancelling the common terms,
λ₁/ λ₂ = n₁²/ n₂² = 3²/ 4² = 9/16
Hence, the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom is 9:16.
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