Question

6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm. B. the electron makes a transition from the n

Answer 1

Complete Question

(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .

(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.

Answer:

A

  $\Delta E = 337 \ eV$

B

  $\lambda = 3.439 *10^{-9} \ m$

Explanation:

Considering question a

From the question we are told that

 The width of the box is  $w = 0.125 \ nm = 0.125 *10^{-9} \ m$

Generally the energy level of a particle confined to a box is mathematically represented as

         $E_n = \frac{n^2 h^2}{8 m L^2 }$

Generally the excitation energy is mathematically represented as

         $\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ]$

From the question  $n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1$  

   Here h  is the Planck's constant with a value  of  $h = 6.62607015 * 10^{-34} J \cdot s$

         m is the mass of  electron with value  $m = 9.11 *10^{-31} \ kg$

So

        $\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ]$

=>     $\Delta E = 539 *10^{-19} \ J$

=>     $\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J$

=>     $\Delta E = 337 \ eV$

Considering question b

Generally the energy level of a particle confined to a box is mathematically represented as

         $E_n = \frac{n^2 h^2}{8 m L^2 }$

Generally the excitation energy is mathematically represented as

         $\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ]$

From the question  $n_2 = 4 \ \ and \ \ n_1 = 1$

 Here h  is the Planck's constant with a value  of  $h = 6.62607015 * 10^{-34} J \cdot s$

         m is the mass of  electron with value  $m = 9.11 *10^{-31} \ kg$

So

        $\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ]$

=>      $\Delta E = 578 *10^{-19} \ J$

=>      $\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}$  

=>      $\Delta E = 361.45 \ eV$  

Gnerally the wavelength is mathematically represented as

          $\lambda = \frac{hc}{\Delta E }$

=>       $\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }$

=>       $\lambda = 3.439 *10^{-9} \ m$