Question

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. 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

Answer 1

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

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

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

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

              $n_1 = 5$