Obtain an expression for the energy levels (in MeV) of a neutron confined to a one dimensional box
1.00x10 -14 m wide.
What is the neutrons minimum energy?
(The diameter of atomic nucleus is of this order of
magnitude.)
Answers
neutrons minimum energy is 20.5 MeV
It is given that a neutron confined to a one dimensional box 1.00 × 10^-14 m wide.
we have to find neutrons minimum energy.
En = n²h²/8mL²
= n² [h²/8mL²]
= n² [ (6.63 × 10^-34 Js)²/8 × (1.67 × 10^-27 kg)(10^-14 m)²]
= n² [43.9569 × 10^-68/(13.36 × 10^(-27 - 28)]
= n² [43.9569 × 10^-68/13.36 × 10^-55]
= n² [3.29 × 10^-13 J]
we know, 1eV = 1.6 × 10^-19 J
so, 3.29 × 10^-13 J = 3.29 × 10^-13/(1.6 × 10^-19)
= 2.05 × 10^6 eV
= 2.05 MeV
so, En = n² × 20.5 MeV
to get minimum energy, n = 1
so, E₁ = 1² × 20.5 MeV = 20.5 MeV
therefore, neutrons minimum energy is 20.5 MeV
Answer:
we have to find neutrons minimum energy.
En = n²h²/8mL²
= n² [h²/8mL²]
= n² [ (6.63 × 10^-34 Js)²/8 × (1.67 × 10^-27 kg)(10^-14 m)²]
= n² [43.9569 × 10^-68/(13.36 × 10^(-27 - 28)]
= n² [43.9569 × 10^-68/13.36 × 10^-55]
= n² [3.29 × 10^-13 J]
we know, 1eV = 1.6 × 10^-19 J
so, 3.29 × 10^-13 J = 3.29 × 10^-13/(1.6 × 10^-19)
= 2.05 × 10^6 eV
= 2.05 MeV
so, En = n² × 20.5 MeV
to get minimum energy, n = 1
so, E₁ = 1² × 20.5 MeV = 20.5 MeV