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Masja [62]
3 years ago
9

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave

length 3745 nm. What is the final level of the electron?(c=3.00×10^8m/s, h=6.63×10^-34 J·s, RH=2.179×106-18J)a. 5
b. 6



c. 8



d. 9



e. 1
Chemistry
1 answer:
timofeeve [1]3 years ago
4 0

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

n_i=8

\lambda=3745\ nm

The amount of energy change during the transition is given by :

\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

And

\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

Plugging all the values we get :

\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5

So, the final level of the electron is 5.

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