Chapter 8: Review

The chapter problem, solved

The hospital prepares a fluorine-18 tracer for a scan. Its half-life is 110 minutes. At the start of the morning its activity is $8.0 \times 10^{8}\,\text{Bq}$ , and the scan needs at least $2.0 \times 10^{8}\,\text{Bq}$ . Each positron annihilates with an electron and produces two 511 keV gamma photons. The detector photocathode has a work function of 2.00 eV, and the team tests it with 250 nm ultraviolet light. The teaching electron gun runs at 150 V.

1. The energy and frequency of a gamma photon. The photon energy is 511 keV.

\begin{aligned} E &= 511 \times 10^{3} \times 1.60 \times 10^{-19} = 8.18 \times 10^{-14}\,\text{J} \\ f &= \frac{E}{h} = \frac{8.18 \times 10^{-14}}{6.63 \times 10^{-34}} = 1.23 \times 10^{20}\,\text{Hz} \end{aligned}

Its wavelength is $\lambda = \dfrac{c}{f} = 2.43 \times 10^{-12}\,\text{m}$ .

2. The ultraviolet test. The photon energy of 250 nm light is

E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{250 \times 10^{-9}} = 7.96 \times 10^{-19}\,\text{J} = 4.97\,\text{eV}

This is above the 2.00 eV work function, so electrons are released. Their maximum kinetic energy is

E_{k(\text{max})} = hf - \phi = 4.97 - 2.00 = 2.97\,\text{eV}

The stopping potential is 2.97 V. The threshold wavelength is 622 nm, so the 250 nm light is well above the threshold and a red lamp would not release electrons.

3. The electron gun. An electron accelerated through 150 V gains

E_k = eV = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17}\,\text{J}

Its momentum is

p = \sqrt{2mE_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-17}} = 6.61 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}

and its de Broglie wavelength is

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10}\,\text{m}

4. The decay constant of fluorine-18. The half-life is $110 \times 60 = 6600\,\text{s}$ , so

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6600} = 1.05 \times 10^{-4}\,\text{s}^{-1}

5. The activity through the morning. 330 minutes is three half-lives.

A = \frac{8.0 \times 10^{8}}{2^{3}} = 1.0 \times 10^{8}\,\text{Bq}

550 minutes is five half-lives, so $A = \dfrac{8.0 \times 10^{8}}{2^{5}} = 2.5 \times 10^{7}\,\text{Bq}$ .

6. The number of undecayed nuclei at the start.

N = \frac{A}{\lambda} = \frac{8.0 \times 10^{8}}{1.05 \times 10^{-4}} = 7.6 \times 10^{12}

7. The energy from the annihilation. The electron and the positron together have a rest energy of $2 \times 0.511 = 1.022\,\text{MeV}$ .

\begin{aligned} E &= 1.022 \times 1.60 \times 10^{-13} = 1.64 \times 10^{-13}\,\text{J} \\ E_{\text{photon}} &= \frac{1.64 \times 10^{-13}}{2} = 8.18 \times 10^{-14}\,\text{J} \end{aligned}

Each photon carries 0.511 MeV. This matches the gamma photons detected by the scanner, so the 511 keV photons really do come from the annihilation.

8. The judgement on the tracer. The activity must be at least $2.0 \times 10^{8}\,\text{Bq}$ . This is one quarter of the starting activity, so it is reached after two half-lives, or 220 minutes. At 220 minutes the tracer is at its weakest usable level. At 330 minutes the activity has fallen to $1.0 \times 10^{8}\,\text{Bq}$ , which is too low. The team should start the scan within 220 minutes of preparing the tracer, or prepare a fresh sample. After that, a longer scan would deliver too small a signal.

9. The first prediction. The activity after 330 minutes is not a half, a quarter or a thousandth. It is one eighth, because 330 minutes contains three intervals of 110 minutes and each interval halves the activity.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate. Some questions bring back ideas from earlier chapters.

A photon has an energy of 2.5 eV. Find its frequency.
Light of wavelength 450 nm falls on a metal of work function 2.20 eV. Find the maximum kinetic energy of the electrons, in eV.
An electron moves at 5.0e6 m/s. Find its de Broglie wavelength.
An electron is accelerated through 250 V. Find its de Broglie wavelength.
A source has an activity of 1200 Bq and a half-life of 8.0 days. Find its activity after 32 days.
A nuclide has a half-life of 8.0 days. Find its decay constant in s^-1.
A nucleus has a mass defect of 0.025 u. Find its binding energy in MeV.
An electron and a positron annihilate. Find the total energy released, in joules.
Write 250 nm in metres.
A laser of wavelength 633 nm passes through two slits 0.20 mm apart, and the screen is 2.0 m away. Find the fringe spacing.
An electron is accelerated from rest through 400 V. Find its speed.
A half-life is measured as 5.35 minutes with an uncertainty of 2.0%. Find the absolute uncertainty.
A source falls from 800 Bq to 100 Bq in 24 hours. Find its half-life.
A beam of electrons is accelerated through 300 V. State which equation you would use to find their de Broglie wavelength, and why.
State which equation you would use to find the energy released when a nucleus loses mass, and why.

Answers and concise solutions

1. The frequency

Answer: $f = \dfrac{E}{h} = \dfrac{2.5 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.0 \times 10^{14}\,\text{Hz}$ .

2. The maximum kinetic energy

Answer: The photon energy is $\dfrac{hc}{\lambda} = 4.42 \times 10^{-19}\,\text{J} = 2.76\,\text{eV}$ , so $E_{k(\text{max})} = 2.76 - 2.20 = 0.56\,\text{eV}$ .

3. The de Broglie wavelength

Answer: $\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 5.0 \times 10^{6}} = 1.5 \times 10^{-10}\,\text{m}$ .

4. The de Broglie wavelength

Answer: $E_k = 4.0 \times 10^{-17}\,\text{J}$ , then $p = \sqrt{2mE_k} = 8.54 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}$ , so $\lambda = 7.8 \times 10^{-11}\,\text{m}$ .

5. The activity

Answer: 32 days is four half-lives, so $A = \dfrac{1200}{2^4} = 75\,\text{Bq}$ .

6. The decay constant

Answer: $T_{1/2} = 8.0 \times 86400 = 6.91 \times 10^{5}\,\text{s}$ , so $\lambda = \dfrac{\ln 2}{T_{1/2}} = 1.0 \times 10^{-6}\,\text{s}^{-1}$ .

7. The binding energy

Answer: $E = 0.025 \times 931.5 = 23.3\,\text{MeV}$ .

8. The total energy

Answer: $E = 2 \times 0.511\,\text{MeV} = 1.64 \times 10^{-13}\,\text{J}$ .

9. The wavelength

Answer: $250\,\text{nm} = 2.50 \times 10^{-7}\,\text{m}$ .

10. The fringe spacing

Answer: From Section 6.1, $w = \dfrac{\lambda D}{s} = \dfrac{633 \times 10^{-9} \times 2.0}{0.20 \times 10^{-3}} = 6.3 \times 10^{-3}\,\text{m}$ .

11. The speed

Answer: The kinetic energy is $E_k = eV = 6.4 \times 10^{-17}\,\text{J}$ . Then $\dfrac{1}{2}mv^2 = E_k$ , so $v = \sqrt{\dfrac{2E_k}{m}} = 1.2 \times 10^{7}\,\text{m}\,\text{s}^{-1}$ .

12. The absolute uncertainty

Answer: $0.020 \times 5.35 = 0.107\,\text{minutes}$ , so the half-life is $5.35 \pm 0.11\,\text{min}$ .

13. The half-life

Answer: 800 to 100 is three halvings, so the half-life is $\dfrac{24}{3} = 8.0\,\text{hours}$ .

14. A beam of electrons is accelerated through 300 V. State which equation you would use to find their de Broglie wavelength, and why.

First find the kinetic energy of an electron from $E_k = eV$ .
Then find the momentum from $p = \sqrt{2mE_k}$ .
Then use the de Broglie equation, $\lambda = h/p$ .
The photon equation $E = hf$ does not apply, because an electron is not a photon.

15. State which equation you would use to find the energy released when a nucleus loses mass, and why.

Find the mass defect, the difference between the masses before and after.
Then use $E = mc^2$ with the mass defect to find the energy released.
The photon equation $E = hf$ does not apply unless a photon of a known frequency is emitted.
The method works because the total mass-energy is conserved, so the missing mass appears as energy.

What you can now do

describe light as a stream of photons and use the photon energy equations;
explain the photoelectric effect and use the photoelectric equation;
find a threshold frequency, a threshold wavelength and a stopping potential;
explain why intensity and frequency play different roles;
use the de Broglie equation for an electron and for a large object;
explain how electron diffraction shows that particles have wave properties;
define activity and half-life, and correct a count rate for the background;
find a decay constant and use the exponential decay law;
write balanced nuclear equations for alpha, beta and gamma decay;
explain why radioactive decay is random and spontaneous;
define mass defect and binding energy, and use $E = mc^2$ ;
explain why fission and fusion release energy;
describe annihilation and find the energy of the photons it produces;
judge whether a radioactive tracer is still strong enough to use.

Your relevance reflection

Return to the prompt from the chapter opener: think of a light or a radiation you have met that can affect a material, such as sunlight on a solar cell, ultraviolet light on a fluorescent label, a smoke alarm, or a hospital X-ray. Write one more example now that you have finished the chapter. Has the chapter changed how you think about what light and radiation do to matter?