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What is the wavelength (in nm) of the photon absorbed for a transition of an electron from ninitial = 2 that results in the least energetic spectral line in the visible series of the H atom?

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Answer: 656.6 nm.

Step-by-step explanation:

Using Rydberg's Equation for hydrogen atom:


(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )

Where,


\lambda = Wavelength of radiation


R_H = Rydberg's Constant


n_f = Higher energy level = 3 (least energetic for visible series)


n_i= Lower energy level = 2

We have:


n_f=3, n_i=2


R_H=1.097* 10^7 m^(-1)


(1)/(\lambda)=1.097* 10^7 m^(-1)* \left((1)/(2^2)-(1)/(3^2) \right )


(1)/(\lambda)=1.097* 10^7 m^(-1)* (5)/(36)


(1)/(\lambda)=0.1523* 10^(7) m


\lambda=6.566* 10^(-7)m=656.6nm

(
1 m= 10^9nm)

The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.

User Ben Brandt
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