Question

Electromagnetic radiation is emitted when a neutron and proton fuse to make a deuteron. What is the wavelength of this radiation?

Answer 1

To arrive at the solution of this question, first we need to find binding energy of deuteron and then find the wavelength of the emitting radiation.

So, for calculating ^2​D binding energy:

Number of protons = 1

Calculated mass of proton, m_p: ​= 1.007276466879 amu

Number of neutrons = 1

Calculated mass of neutron, m_n: = 1.008664915885 amu

Calculated total mass of proton and neutron (m_n+m_p) = 2.015941382764 ​amu

Observed deuteron mass, m_​o = 2.013553212745 amu

Since mass defect, M_d

M_d = (m_n+m_p)−m_o

_​                                         M_d    = 2.015941382764 - 2.013553212745

                                    = 0.002388170019 amu

Convert amu into Kg units

                                     = 0.002388170019 amu x  1.660539040 x 10^-27 kg/amu

                    ​                       = 0.003965649550707 x 10^​-27​ kg

Now calculate the energy:

E = mc²

E = 0.003965649550707 ​x 10^​-27​ kg x (2.99792458 x 10⁸ m/s )²

    = 0.003965649550707 ​x 10^​-27​ kg x 8.98755179 x 10^​16 ​

Binding energy of ²D nucleus, E  = 0.0356414807179698 x 10^-11​ J/atom

Now calculate wavelength of emitting radiation:

Wavelength = Planck’s constant x speed of light / energy

Planck constant, h = 6.62607004081818 x10^-34 J s

Wavelength = 6.62607004081818 x10^-34 x 2.99792458 x 10⁸ ​/ 0.0356414807179698 x 10^-11​

       = 19.86445824417043 x 10^-26 / 0.0356414807179698 x 10^-11​

       = 19.86445824417043 x 10^-15 / 0.0356414807179698

       = 557.3409926865109 x ​10^-15

       = 557.3409926865109 x ​10^-15m / 1 x 10^-9

       = 5.57340993 x ​10^-4 nm

Wavelength of radiation = 5.57340993 x ​10^-4 nm = 0.000557340993 nM

Gamma rays region = 0.01 – 0.0001 nM.

The wavelength of electromagnetic radiation emitted is within the limits of gamma rays region        

and hence it is a gamma radiation.