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1 Answers

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