Question

Consider an isotope with an atomic number of [2(5+A)] and a mass number of [4(5+A)+2]. Using the atomic masses given in the attached table, calculate the binding energy per nucleon for this isotope. Give your answer in MeV/nucleon and with 4 significant figures. (A=8)

A particular radioactive isotope has a half-life of (2.50+A) hours. If you have (24.5+B) g of the isotope at 10:00 AM, how much will you have at 7:30 PM? Give your answer in grams (g)and with 3 significant figures. (A=8, B=7)

Answer 1

Solution:

Answers: a) 8.732 MeV/nucleon

b) 19.2 grams

a) ( "Using the atomic masses given in the attached table" is said but Attached table is not given; general values are taken for the masses) .element's atomic no = Z = 2(5+A) = 2(5+8) = 26

Atomic mass = A = 4 (5+A) = 4 ( 5+8) +2 = 54

This is an Isotope of Iron . (26 Fe 54)

Mass of proton = 1.0072766 a.m.u. = m_p

Mass of neutron = m_n = 1.0086654 a.m.u.

Mass of the nucleus = m_p( No. of proton) + m_n (No. of neutrons) = 26 ( 1.0072766) + 28 (1.00866654)

= 54.43185 amu

Atomic Mass of Fe 54 = 54.938044

Mass defect = 54.938044 - 54.43185 = 0.506189 amu

1 amu = 931.5 MeV

Binding energy = mass defect x 931.5 = 471.5 MeV

Binding energy per nucleon = 471.5 / 54 = 8.732 MeV /nucleon.

b) Half life = T1/2 = 2.5+A = 2.5+8 = =10.5 hours

Ao = Amount = 24.5 + B = 24.5 + 7 = 31.5 grams

Final amount = A = Ao (1/2) ^{t /T1/2}

= (31.5) (1/2) ^{7.5 / 10.5} = 19.2 grams