Nuclear Physics

Question 1

decay constant

Answer 1

λ - the probability of an individual nucleus decaying per second

Question 2

unit for the decay constant

Answer 2

s^-1

Question 3

A piece of wood has 45% less C-14 compared to living wood sample with λ = 3.8 * 10^-12 /s. Calculate its age in years.

Answer 3

m = m_{0e^-λt
Let m0 = 100g, then m = 55g
55 = 100 * e^{-(3.8 * 10^-12)t}}

e^{-(3.8 * 10^-12)t} = 11/20

-(3.8 * 10^-12) t = -0.597…
t = 1.573… * 10^11s = 4988.75… ~ 5000 years

Question 4

What is the mass measured in when you’re trying to calculate moles?

Answer 4

grams

Question 5

Why does gamma radiation have the greatest range?

Answer 5

Due to inverse square law

Question 6

Write an equation for power in terms of the activity of a radioactive sample.

Answer 6

Power = energy transfer / time = activity * energy of each particle

Question 7

On a N-Z graph, why do isotopes stop following the N = Z line after Z = 20?

Answer 7

After Z = 20, the nuclei must have more neutrons to remain stable - this increases the size of the strong force as if there are lots of protons = high repulsive electrostatic force.

Question 8

Where are alpha emitters found on a N-Z graph?

Answer 8

Alpha emitters are found above the N = Z line and around (usually below) the stable nuclei curve as they have lots of neutrons (more than protons)
(losing 2 (p + n) therefore they move ↙)

Question 9

Where are beta minus emitters found on a N-Z graph?

Answer 9

Beta minus emitters are found to the left of the stable nuclei line but lower than alpha emitters as they have too many neutrons but not enough protons
(n -> p therefore they move ↘)

Question 10

Where are beta minus emitters found on a N-Z graph?

Answer 10

Beta plus emitters are found to the right of the stable nuclei line as they have too many protons but not enough neutrons
(p -> n therefore they move ↙)

Question 11

metastable isotope

Answer 11

An isotope nucleus that has the correct number of protons & neutrons but too much energy (is excited)

Question 12

How do metastable isotopes release their extra energy? Why do they do this?

Answer 12

Via gamma radiation - deexcites the isotope that that it reaches its ground state.

Question 13

The density of nuclei is ____. Derive an expression and explain your answer.

Answer 13

always the same
Density = m / V = (Au) / [4/3 π (r₀)³A] = u / [4/3 π (r₀)³ ] = 3u / [4 π (r₀)³]
and this expression for density is just filled with constants.

Question 14

Why is the atomic mass unit, u, not equal to the mass of a neutron/proton?

Answer 14

When staying together as a stable carbon 12 nucleus, the six protons and neutrons form a state with slightly lower mass/energy compared to if they were independent particles (due to the binding energy which holds the nucleus together).

Question 15

How can we get a more accurate value for the radius of a nucleus than from the gold foil alpha particle scattering experiment?

Answer 15

High energy electron diffraction - when v is high (~ 3 * 10⁸) the electron has a small enough de Broglie wavelength to diffract through a single slit.

Question 16

Why are there not that many maxima rings in high energy electron diffraction?

Answer 16

There’s attraction between the electron & protons.

Question 17

Formula for the radius of a nucleus in terms of high energy electron diffraction. Name the symbols.

Not in FB

Answer 17

Rsin(θ_min) = 0.61λ
{or, as in textbook, sin(θ_min) = 0.61λ / R}
where R = radius of nucleus
θ = angle between electron beam to first minimum
λ = electron de Broglie wavelength

Question 18

Explain why the first minimum is used to find the radius of a nucleus from high energy electron diffraction.

Answer 18

It is the most accurate - the electrons can be attracted to the protons so can be scattered at higher minima; the higher the order, the harder it is to identify the local minimum.

Question 19

R = r_{0A^1/3. What is the value for A?}

Answer 19

mass (nucleon) number of radioactive sample

Question 20

mass defect

Answer 20

The loss of mass from a nuclear decay - the total mass of a nucleus is less than the total mass of the individual separated nucleons

Question 21

Power output for a nuclear power plant = 6 * 10^{8W, efficiency = 35%
Calculate the decrease in the mass of the fuel rods due to the release of energy for one week of continuous operation.}

Answer 21

Let E_out be useful electrical energy output of the power plant and E_nuc be the total nuclear energy released from the decays inside the reactor core.
Therefore E_out = E_nuc * 0.35

t = 1 week = 604,800s

E_out = (6 * 10⁸) * 604,800 = 3.6288 * 10¹⁴J
E_nuc = 3.6288 * 10¹⁴ / 0.35 = 1.0368 * 10¹⁵J

OR in terms of power:
P_out = 6 * 10⁸ / 0.35 = 1.714… * 10⁹ W (P_out = the 35% that is released usefully)
ΔE =1.714… * 10⁹ 604,800 = 10¹⁵J

ΔE = mc² -> 1.0368 * 10¹⁵ = m (3 * 10⁸)²
Therefore the mass decrease = 0.01152 ~ 1.2 * 10^-2kg

Question 22

Why does alpha scattering not give an accurate value for nuclear radius? Will it always be an overestimate or an underestimate?

Answer 22

This is where you use Coulomb’s law. Ep gold nucleus = Ek alpha particle, then rearrange for r.
Overestimate, because it measures the smallest separation between the alpha particle and the gold nucleus, not the nucleus’ radius

Question 23

In beta minus decay, what energy(-ies) does the beta particle emitted have? Exxplain your answer.

Answer 23

Antineutrino released has a range of energies, therefore the beta particle has a range of energies.

Question 24

The speed of all the electrons in high energy electron diffraction need to be kept constant. Explain why.

Answer 24

To maintain the SAME de Broglie wavelength which is similar in size to the nucleus, in order for the electrons to diffract the same amount.
If there is an excessive amount of scattering, then the first minimum can be difficult to determine.