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Alenkasestr [34]
2 years ago
11

A hydrogen atom in an excited state emits a photon of frequency ν = 3.084 x 1015 s-1. If the electron returns to the ground stat

e, in which level was it before the photon was emitted?
Chemistry
1 answer:
VLD [36.1K]2 years ago
4 0

<u>Answer:</u> The photon was emitted from the 4th energy level

<u>Explanation:</u>

To calculate the wavelength of light, we use the equation:

\lambda=\frac{c}{\nu}

where,

\lambda = wavelength of the light  

c = speed of light = 3\times 10^8m/s

\nu = frequency of light = 3.084\times 10^{15}s^{-1}

Putting the values in above equation, we get:

\lambda=\frac{3\times 10^8m/s}{3.084\times 10^{15}s^{-1}}=9.73\times 10^{-8}m

To calculate the initial level, we use the equation given by Rydberg's equation, which is:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )

where,

\lambda = Wavelength of radiation  9.73\times 10^{-8}m

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Final energy level =  1  

n_i= Initial energy level =  ?

Putting the values, in above equation, we get

\frac{1}{9.73\times 10^{-8}m}=1.097\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{n_i^2} \right )

n_i=4

Hence, the photon was emitted from the 4th energy level

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