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ikadub [295]
3 years ago
5

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 x 108m/s, h = 6.63 x 10−34 J • s, RH = 2.179 x 10−18 J)
a. 5
b. 6
c. 8
d. 9
e. 1
Chemistry
1 answer:
topjm [15]3 years ago
7 0

Answer:

The level is  n_1 = 5

Explanation:

From the question we are told that

  The level of the hydrogen atom is   n_2 =  8

  The wavelength of the photon is  \lambda =  3745 nm =  3745 *10^{-9} \ m

Generally the wave number is mathematically represented as

        k  =  \frac{1}{\lambda }

Now this wave number can also be mathematically represented as

       k = R_{\infty} [\frac{1}{n_1^2} +  \frac{1}{n_2^2}  ]  

This implies that

     

 So  

   Here R_{\infty} is the Rydberg constant, with a value  1.097 * 10^7

and  n_1 \ and \ n_2  are the principal quantum levels

 substituting values

               0.0243=  [\frac{1}{n_1^2} - \frac{1}{8^2}  ]  

               0.0243=  \frac{1}{n_1^2} - 0.015625  

              0.0243 + 0.015625=  \frac{1}{n_1^2}

              n_1 = 5  

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