Answer:
First Question
  [tex]E = 1.065*10^{-12} \ J[/tex]
Second  Question
  The  wavelength is for an X-ray Â
Explanation:
From the question we are told that
   The  width of the wall is  [tex]w = 10\ fm = 10*10^{-15 }\ m[/tex]
   The  first excited state is  [tex]n_1 = 2[/tex]
   The  ground state is  [tex]n_0 = 1[/tex]
Gnerally the  energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as
     [tex]E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }[/tex]
Here  h is the Planck's constant with value  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
     m is the mass of proton with value [tex]m = 1.67 * 10^{-27} \ kg[/tex]
So  Â
     [tex]E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }[/tex]
=> Â Â Â Â [tex]E = 1.065*10^{-12} \ J[/tex]
Generally the energy of the photon emitted is also mathematically represented as
       [tex]E = \frac{h * c }{ \lambda }[/tex]
=> Â Â Â Â Â [tex]\lambda = \frac{h * c }{E }[/tex]
=> Â Â Â Â Â [tex]\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }[/tex]
=> Â Â Â Â [tex]\lambda = 1.87*10^{-10} \ m[/tex]
Generally the range of wavelength of X-ray is  [tex]10^{-8} \to 1)^{-12}[/tex]
So this wavelength is for an X-ray.
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The energy emitted is  6.1625 MeV and the spectrum is a gamma-ray spectrum.
The energy of the [tex]n_{th}[/tex] level associated to the infinite square well potential is:
[tex]E_{n} =\frac{n^{2}h^{2} }{8mL^{2} }[/tex]
here, h = Planck's Constant = [tex]6.626*10^{-34} Js[/tex]
     m = mass of proton = [tex]1.67*10^{-27}kg[/tex]
     L = width of the square potential = [tex]10*10^{-15} m[/tex]
according to the question the proton undergoes a transition from the first excited state (n = 2) to the ground state (n = 1)
ΔE = [tex]E_{2}-E_{1}[/tex]
   [tex]=\frac{2^{2}(6.626*10^{-34} )^{2} }{8*1.67*10^{-27}* (10*10^{-15} )^{2} }+\frac{1^{2}(6.626*10^{-34} )^{2} }{8*1.67*10^{-27}* (10*10^{-15} )^{2} }[/tex]
ΔE [tex]= 9.86*10^{-13}J[/tex] = 6.1625 MeV is the energy emitted.
Now,  ΔE = hc/λ
where λ is the wavelength of light emitted , c is the speed of light
λ = hc/ΔΕ
 [tex]=\frac{6.626*10^{-34}*3*10^{8} }{ 9.86*10^{-13}}[/tex]
λ = [tex]2.016*10^{-13} m[/tex]
This wavelength belongs to gamma-ray spectrum.
Learn more about infinite square well :
https://brainly.com/question/13777893?referrer=searchResults
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