Question

6) The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; 1 eV = 1.602 × 10–19 J. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with n = 5 to the orbit with n = 2. Show your calculations.

Answer 1

Answer:

The energy of the photon is  $x = 2.86 eV$

Explanation:

From the question we are told that  

      The first orbit is $n_1 = 5$

       The second orbit is $n_2 = 2$

According to  Bohr model

     The energy of difference of the electron as it moves from on orbital to another is mathematically represented as

              $\Delta E = k [\frac{1}{n^2 _1} + \frac{1}{n^2 _2} ]$

  Where k is a constant which has a value of $k = -2.179 *10^{-18} J$

       So

              $\Delta E = - 2.179 * 10^{-18} [\frac{1}{5^2 _1} + \frac{1}{2^2 _2} ]$

                     $= 4.576 *10^{-19}J$

Now we are told from the question that

         $1 eV = 1.602 * 10^{-19} J$

so      x eV  =  $= 4.576 *10^{-19}J$

  Therefore

                $x = \frac{4.576*10^{-19}}{1.602 *10^{-19}}$

                   $x = 2.86 eV$