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Chapter 11: Q19E (page 518)

Show thatthe energy required to remove a neutron from helium-4 is .20.6MeV

Short Answer

Expert verified

Energy required to remove one neutron from helium- 4is 20.6MeV.

Step by step solution

01

Given data

Mass of neutron, mn=1.008665u.

Atomic mass of helium, −4,M42He=4.002603u.

Speed of light, c=3×108m/s.

02

Formula for Binding energy

For helium- 4, binding energy is given by formula, BE=(ZmH+Nmn−M41x)c2.

03

Calculation for energy required to remove neutron

After removing 1 neutron from helium-4, the mass of helium becomes:

MHe−3=3.016029uand number of neutron becomes.N=1

The binding energy is given as:

BE=(MHe−3+Nmn−M42He)c2BE=[3.016029u+(1)(1.008665u)−4.002603u](3×108m/s)2

The conversion factor may be used to express c2in MeV/u.

BE=[3.016029u+(1.008665u)−4.002603u](931.5MeVu)BE=20.6MeV

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