Answers to questions on nuclear fusion & fission reactions (HL)

Answers

1. (a) Mass = (92 x 1.672622.10⁻²⁷) + (143 x 1.674927.10⁻²⁷) kg

= (153.881224 + 239.514561) x 10^-27 = 3.933958 x 10^-25 kg

(b) Mass of one mole of U-235 nuclei = (3.933958 x 10^-25) x (6.022142 x 10²³) = 2.369085 x 10^-1 kg
Mass defect = (2.369085 – 2.350439) x 10^-1 = 1.8646 x 10^-3 kg mol^-1
Binding energy = 1.8646 x 10^-3 x (2.997925 x 10⁸)² = 1.67582 x 10¹⁴ J mol^-1 = 1.68 x 10¹¹ kJ mol^-1

(c) Nuclear mass of U-235 + one neutron = 235.043924 + (1.674927 x 10^-24 x (6.022142 x 10²³))
= 236.0526 g mol^-1
Mass defect = (236.0526 – 235.8454) = 0.2072 g = 2.072 x 10^-4 kg
Energy released = (2.072 x 10^-4) x (2.997925 x 10⁸)² = 1.86 x 10¹³ J mol^-1
= 1.86 x 10¹⁰ kJ mol^-1

2.(a) The temperature of a gas is proportional to its average kinetic energy, ½ mv² where m is the mass of the gas and v is its average velocity. For two different gases, 1 and 2, at the same temperature

½ m₁v₁² = ½ m₂v₂²

v₁ / v₂ = √ (m₂ / m₁) = √ (M_r2 / M_r1)

i.e. the rate at which gases diffuse is inversely proportional to the square root of their molar masses.

(b) M_r(C₂H₆) = 30.06
v₁ / v₂ = 1.37 = √ (30.06 / M_r)
M_r = 30.06 / 1.37² = 16.02

(c) v₁ / v₂ = √ (28.02 / 2.02) = 3.72
Hydrogen will effuse 3.72 times faster than nitrogen.

3.(a)

(b) The graph shows that it takes the same time for the amount of strontium-90 to be halved irrespective of the initial amount, i.e. the time to go from 100 g to 50 g is the same as the time to go from 50 g to 25 g or from 25 g to 12.5 g.

(c) Five half-lives = 144 years so t_½ is 28.8 years.
λ = ln 2 / 28.8 = 0.0241 years^-1
N = N_oe^−λt = 100 x e^{−(0.0241 x 200)} = 0.807 g

(d) Strontium is immediately below calcium in Group 2 of the periodic table. When strontium-90 is ingested (from water or milk for example) it gets taken up by the body in bones (in place of calcium) and so remains in the body for a relatively long period emitting
beta radiation.

Answers to Nuclear fusion & fission reactions questions