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Please answer this question :-
What is the frequency of a photon emitted during a transition from n=5 state to n=2 state in the hydrogen atom ? Which region of the spectrum does this trasition correspond to ? [ RH= 2.18 X 10^-18 J }
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Answers
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HERE'S UR ANS....
Use Rydberg's formula, Which is,
1/∆ = RH(1/n1^2-n2^2)z^2
∆(lambda) = which is the wavelength of the photon emitted.
Where RH is Rydberg's constant = 1.097 x 10^7 m^-1
n^1 = lower energy state, i.e 2 in this question.
n^2 = higher energy state, i.e 5
Z = atomic number, here Z for hydrogen is 1
On substituting the given values we get,
1/∆ = 1.097 x 10^7(1/2^2–1/5^2)1^2
1/∆ = 1.097x 10^7(1/4–1/25)
1/∆ = 1.097 x 10^7(0.21)
1/∆ = 0.23 x 10^7
On doing reciprocal we get ∆(lambda), i.e the wavelength of the emitted photon.
∆= 4.347 x 10^-7 m
But we require its frequency,
Therefore, frequency = c/∆
(c= speed of light)
3x10^8/4.347x10^-7 = 0.69 x 10^15
Therefore, frequency= 0.69 x 10^15
Anonymous:
my pleasure ;)
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