1. Consider the following nuclides: &

1. Consider the following nuclides:

NUCLIDE MASS (amu)

₅¹¹B 11.009305

₆¹²C 12.0000

₄₂¹⁰⁰Mo 99.90747

₈₃²⁰⁹ Bi 208.97973

₉₂²³⁸ U 238.050

For each of these nuclides, find (a) the binding mass in atomic mass units and (b) the binding energy in Mev.

2. (a) Find the average binding energy per nucleon of each nuclide listed in problem 1, and (b) mark the location of each nuclide on the attached graph of average binding energy per nucleon versus atomic mass number.

3. Find the average radius of each nuclide, listed in problem 1.

Expert Solution

The binding energy is the mass difference between the mass of the nuclide and the sum of the mass of its constituents, expressed in energy by means of Einstein's equation E=mc². The mass of the proton is 1.007276 amu, the mass of the neutron is 1.008665 amu and the mass of the electron is 0.000549 amu. The mass of the nucleus is the measured mass minus the mass of the electrons. The mass of the constituents is the sum of number of nucleons times the mass of the nucleon. If he have the mass defect in amu, we can multiply by 931.3 to convert to MeV. To get the binding energy per nucleon, we divide by the number of nucleons (protons + neutrons). The avg radius of a nucleus with A nucleons is R = R₀A¹^/³, where R₀ = 1.2E^-15 m.

Element	m_nucleus (amu)	m_const (amu)	m_diff (amu)	E(MeV)	E_nucleon(MeV)	R_avg(m)
	11.009305-5x0.000549	5x1.007276+6x1.008665	-0.08181	76.19	6.97	4.4E-15
	12-12x0.000549	6x1.007276+6x1.008665	-0.102234	95.21	7.93	4.8E-15
	99.90747-42x0.000549	42x1.007276+58x1.008665	-0.92375	860.29	8.6029	4E-14
	208.97973-83x0.000549	83x1.007276+126x1.008665	-1.761535	1640.52	7.85	8.36E-14
	238.05-92x0.000549	92x1.007276+146x1.008665	-1.93499	1802.06	7.57	9.52E-14

genius_generous answered 6 hours ago

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