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1. The mass of a proton is 1.00728 amu and that of a neutron is 1.00867 amu. Wha

ID: 1039879 • Letter: 1

Question

1. The mass of a proton is 1.00728 amu and that of a neutron is 1.00867 amu. What is the binding energy (in J) of a 60/27Co nucleus? (The mass of a cobalt-60 nucleus is 59.9338 amu. Speed of light = 3.00 × 108 m/s.)

a. 2.74 × 10^-19

b. 2.74 x 10^-16

c. 8.20 x 10^-11

d. 9.12 x 10^-28

e. 4.94 x 10^-13

2. What percentage of a radioactive sample remains after 175.0 yr if it has a half-life of 28.8 years?

a. 6.08

b. 84.8

c. 1.48

d. 0.230

e. 16.5

3. The half-life of carbon-11 is 20.3 minutes. How much of a 100.0 mg sample remains after 1.6 hours?

a. 99.9 mg

b. 1.75 x 10^-89 mg

c. 94.7 mg

d. 0.0377 mg

4. The half-life of cobalt-60 is 5.20 yr. How many milligrams of a 2.000-mg sample remain after 9.50 years?

a. 7.076

b. 0.565

c. 1.435

d. 7.03 x 10^-22

e. 1.095

e. 3.77 mg

Explanation / Answer

1.

binding energy(E) = DmC^2

Dm = mass defect

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

Expected mass of Co = (27*1.00728+33*1.00867) = 60.5 amu

Actual mass of Co = 60 amu

Dm = Expected mass of Ba - Actual mass of Ba

    = 60.5-60

    = 0.5 amu

    = 0.5/(6.022*10^26) kg        (1 amu = 1/6.022*10^26 kg)

= 8.3*10^-28 kg

E = DmC^2

= (8.3*10^-28)*(3*10^8)^2

= 7.47*10^-11 joule

answer: C

2.

first order kinetics

K1 = 0.693/t1/2

   = 0.693/28.8

   = 0.024 days-1


k1 = (1/t)ln(a0/a)

a0 = initial concentration = 100

a = concentration remains after 175 years = x

t = time = 175 years

0.024 = (1/175)ln(100/x)

x = 1.48%

answer: C

3. K1 = 0.693/t1/2

   = 0.693/20.3

   = 0.034 min-1


k1 = (1/t)ln(a0/a)

a0 = initial concentration = 100

a = concentration remains after 1.6 hrs = x

t = time = 1.6*60 = 96 min

0.034 = (1/96)ln(100/x)

x = 3.82 mg

no answer available

4. K1 = 0.693/t1/2

   = 0.693/5.2

   = 0.133 y-1


k1 = (1/t)ln(a0/a)

a0 = initial concentration = 100

a = concentration remains after 9.5 years = x

t = time = 1.6*60 = 96 min

0.133 = (1/9.5)ln(2/x)

x = 0.565

answer: b.0.565 mg