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AnswerCategory: ChemistryUsing the Rydberg Equation
Petter Smith Staff asked 3 weeks ago

Using the Rydberg Equation

Using the Rydberg Equation, how do I calculate the wavelength (in nanometers) of the light emitted when the electron in a hydrogen atom undergoes a transition from n = 6 to n =2?

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1 Answers
Petter Smith Staff answered 3 weeks ago

Energy of a photon emitted when an electron returns back from its higher energy state to the lower energy states is given by the formula,
E = RE ( 1/n1^2 – 1/n2^2 )
Here RE stands for Rydberg constant which is expressed in Joules, n1 is the final energy state and n2 is the initial energy state. In this case, n1 = 2 and n2 = 6.
Thus, E = 2.18 * 10^-18 J (1/2^2 – 1/6^2 )
= 2.18 * 10^-18 J (1/4 – 1/36 )
= 4.84 * 10^-19 J
According to Plank – Einstein Equation,
E = h * ϑ, where
E = energy of photon,
h = Plank’s constant, that is, 6.626 * 10^-34 J s
ϑ = frequency of photon
We know,
ϑ * λ = c
where, λ is the wavelength and ‘c’ is the speed of light in vacuum which is equal to 3.0 * 10^8 ms^-1 approx.
Thus, the relationship between energy and wavelength will be:
E = h * c / λ
Incorporating the definite values, we get,
λ = ( h * c ) / E
= {(6.626 * 10^-34 J s) * (3.0 * 10^8 m s^-1)} / 4.84 * 10^-19 J
λ = 4.11 * 10^-7 m
or, λ = 411 nm