Question

A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by an infinite square well potential). Calculate the energy (in MeV) of the photon emitted when the proton undergoes a transition from the first excited state (n = 1) to the ground state (n = 1). In what region of the electromagnetic spectrum does this wavelength belong?

Answer 1

Answer:

First Question

    $E = 1.065*10^{-12} \ J$

Second  Question

   The  wavelength is for an X-ray  

Explanation:

From the question we are told that

     The  width of the wall is  $w = 10\ fm = 10*10^{-15 }\ m$

     The  first excited state is  $n_1 = 2$

     The  ground state is   $n_0 = 1$

Gnerally the  energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

          $E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }$

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

         m is the mass of proton with value $m = 1.67 * 10^{-27} \ kg$

So    

          $E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }$

=>        $E = 1.065*10^{-12} \ J$

Generally the energy of the photon emitted is also mathematically represented as

             $E = \frac{h * c }{ \lambda }$

=>          $\lambda = \frac{h * c }{E }$

=>          $\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }$

=>         $\lambda = 1.87*10^{-10} \ m$

Generally the range of wavelength of X-ray is  $10^{-8} \to 1)^{-12}$

So this wavelength is for an X-ray.