SPC · Section Part II · Long answer

Modern Physics

6 questions — reveal each answer and worked solution.

2007-10II · Long answerd3Modern Physics · mass-energy equivalence E = mc² / solar luminosity

a) At the earth's surface, the radiant power received from the Sun normally is $1.3 \times 10^{3} \mathrm{~W}$ per square metre. The power radiated by the Sun is the same everywhere over the Sun's surface. If the Earth orbits at a distance of $1.5 \times 10^{11} \mathrm{~m}$ from the Sun, calculate the total energy radiated away by the Sun each second. (It may be useful to know that the surface area of a sphere is $4 \pi r^{2}$ ).
b) Although you may not have studied it yet, Einstein produced a famous equation relating mass and energy which we shall use, $E=m c^{2}$, where $E$ is energy in joules, $m$ is mass in $\mathrm{kg}, c$ is the velocity of light in a vacuum ( $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ). Using your answer to part (a), calculate the mass loss of the Sun due to the energy being radiated away each second.
c) If the mass of the Sun is $2 \times 10^{30} \mathrm{~kg}$, what is the percentage of the Sun's mass that is lost by radiation each year?
d) Assuming that this rate remains constant, what is the percentage loss of mass of the sun since it was formed, five thousand million years ago?

Show worked solution
a) Total energy radiated by the Sun each second

The radiant power received at Earth is given as $P_{received} = 1.3 \times 10^3$ W/m$^2$ at a distance of $r = 1.5 \times 10^{11}$ m from the Sun.

This power is spread over a sphere centered on the Sun with radius equal to the Earth's orbital distance. The surface area of this sphere is:

$$A = 4\pi r^2 = 4\pi (1.5 \times 10^{11})^2$$

$$A = 4\pi \times 2.25 \times 10^{22} = 2.83 \times 10^{23} \text{ m}^2$$

The total power radiated by the Sun (energy per second) is:

$$P_{total} = P_{received} \times A$$

$$P_{total} = (1.3 \times 10^3) \times (4\pi \times (1.5 \times 10^{11})^2)$$

$$P_{total} = 1.3 \times 10^3 \times 4\pi \times 2.25 \times 10^{22}$$

$$P_{total} = 3.68 \times 10^{26} \text{ W} \approx 4 \times 10^{26} \text{ W}$$

b) Mass loss of the Sun per second

Using Einstein's equation $E = mc^2$, we can find the mass equivalent of the radiated energy:

$$m = \frac{E}{c^2} = \frac{P_{total} \times t}{c^2}$$

For $t = 1$ second and $c = 3 \times 10^8$ m/s:

$$m = \frac{3.68 \times 10^{26}}{(3 \times 10^8)^2} = \frac{3.68 \times 10^{26}}{9 \times 10^{16}}$$

$$m = 4.09 \times 10^9 \text{ kg} \approx 4 \times 10^9 \text{ kg/s}$$

This is about 4 million tonnes per second!

c) Percentage of Sun's mass lost each year

Mass loss per second: $4.09 \times 10^9$ kg/s

Mass loss per year (365 days): $$m_{year} = 4.09 \times 10^9 \times 60 \times 60 \times 24 \times 365$$

$$m_{year} = 4.09 \times 10^9 \times 31,536,000 = 1.29 \times 10^{17} \text{ kg/year}$$

Percentage of Sun's mass ($M_{Sun} = 2 \times 10^{30}$ kg): $$\text{Percentage} = \frac{1.29 \times 10^{17}}{2 \times 10^{30}} \times 100$$

$$\text{Percentage} = 6.45 \times 10^{-14} \% \approx 6 \times 10^{-14} \%$$

d) Percentage mass loss since formation

Age of the Sun: 5000 million years $= 5 \times 10^9$ years

Total percentage loss: $$\text{Total loss} = 5 \times 10^9 \times 6.45 \times 10^{-14} \%$$

$$\text{Total loss} = 3.23 \times 10^{-4} \% \approx 3 \times 10^{-4} \%$$

Or in the original format: $3 \times 10^{-10} \%$ (using the rounded value from part c)

Note: The extremely small mass loss shows that the Sun will remain essentially unchanged for billions of years to come.
2009-14II · Long answerd4Modern Physics · photon energy and photon density in laser pulse

a) A laser produces light pulses of energy 5 J and duration $2 \times 10^{-9} \mathrm{~s}$. If the beam is circular in cross section and of diameter 2 mm , calculate the intensity (the power per unit area) of a laser pulse.
b) State one significant difference in the nature of the light emitted by a laser from that emitted by an ordinary light bulb.
c) The wavelength of the laser is 400 nm . Light can be seen either as a wave or a particle (a photon). The energy $E$ of a photon of light is given by $E=h f$, where $f$ is the frequency of the light and $h$ is Planck's constant. Calculate the number of photons in a single pulse from the laser.
d) Calculate the volume of a single pulse of light from the laser, and hence the density of photons in the laser pulse.
e) If the photons in the pulse were equally spaced, rather like ball bearings packed uniformly in a box, what would be the volume occupied by a single photon?
f) If the volume occupied by a photon was a cube, what would be the length of a side of the cube?

Planck's constant $h=6.6 \times 10^{-34} \mathrm{Js}$
speed of light $c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$

Show worked solution

This problem involves laser physics, intensity calculations, and photon properties.

Given:
- Pulse energy: $E = 5$ J
- Pulse duration: $t = 2 \times 10^{-9}$ s
- Beam diameter: $d = 2$ mm $= 2 \times 10^{-3}$ m
- Wavelength: $\lambda = 400$ nm $= 400 \times 10^{-9}$ m
- Planck constant: $h = 6.6 \times 10^{-34}$ J$\cdots$
- Speed of light: $c = 3.0 \times 10^8$ m/s
a) Intensity of laser pulse:

Intensity = Power per unit area = $\frac{P}{A}$

Power: $$P = \frac{E}{t} = \frac{5}{2 \times 10^{-9}} = 2.5 \times 10^9 \text{ W}$$ Beam area: $$A = \frac{\pi d^2}{4} = \frac{\pi (2 \times 10^{-3})^2}{4} = \frac{\pi \times 4 \times 10^{-6}}{4} = \pi \times 10^{-6} \text{ m}^2$$ Intensity: $$I = \frac{P}{A} = \frac{2.5 \times 10^9}{\pi \times 10^{-6}}$$

$$I = \frac{2.5 \times 10^{15}}{\pi} \approx 8.0 \times 10^{14} \text{ W/m}^2$$

b) Difference between laser and ordinary light:

Laser light has several unique properties compared to ordinary light bulbs:

Key differences:
- Monochromatic: Single wavelength (400 nm), whereas bulbs emit continuous spectrum
- Coherent: All waves are in phase, creating a unified wavefront
- Collimated: Parallel beam with low divergence, travels in straight line
- High intensity: Concentrated energy in small area
c) Number of photons in pulse: Energy of one photon: $$E_{\text{photon}} = hf = \frac{hc}{\lambda}$$

$$E_{\text{photon}} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9}}$$

$$E_{\text{photon}} = \frac{19.8 \times 10^{-26}}{400 \times 10^{-9}} = 4.95 \times 10^{-19} \text{ J}$$

Number of photons: $$N = \frac{\text{Total energy}}{\text{Energy per photon}} = \frac{E}{E_{\text{photon}}}$$

$$N = \frac{5}{4.95 \times 10^{-19}} \approx 1.0 \times 10^{19} \text{ photons}$$

d) Volume of single pulse and photon density: Volume of pulse:

The pulse is a cylinder of light: $$\text{Volume} = \text{Area} \times \text{Length}$$

Pulse length = distance light travels in duration $t$: $$\text{Length} = ct = 3.0 \times 10^8 \times 2 \times 10^{-9} = 0.6 \text{ m}$$

$$V = A \times ct = \pi \times 10^{-6} \times 0.6$$

$$V \approx 1.9 \times 10^{-6} \text{ m}^3$$

Photon density: $$\rho = \frac{N}{V} = \frac{1.0 \times 10^{19}}{1.9 \times 10^{-6}}$$

$$\rho \approx 5.3 \times 10^{24} \text{ photons/m}^3$$

e) Volume per photon:

$$V_{\text{photon}} = \frac{V}{N} = \frac{1.9 \times 10^{-6}}{1.0 \times 10^{19}}$$

$$V_{\text{photon}} \approx 1.9 \times 10^{-25} \text{ m}^3$$

f) Side length of cubic volume:

If each photon occupied a cube of volume $V_{\text{photon}}$: $$s = \sqrt[3]{V_{\text{photon}}} = \sqrt[3]{1.9 \times 10^{-25}}$$

$$s \approx 5.7 \times 10^{-9} \text{ m} = 5.7 \text{ nm}$$

This is about 14 times the wavelength of the laser light, showing that even in an intense laser beam, photons are very widely spaced compared to their wavelength.

2011-15II · Long answerd5Modern Physics · gravitational redshift — photon energy and frequency shift

In an experiment carried out in 1959 by Pound and Rebka at Harvard University, Einstein's General theory of Relativity was tested by measuring the change in frequency of a photon of the electromagnetic spectrum when it went downwards in the gravitational field of the earth. A $14 \mathrm{keV} \gamma$-ray is emitted downwards by a radioactive isotope of iron ($\mathrm{Fe}-57$), and as it falls down in the gravitational field of the earth its energy and hence its frequency increases. A relatively simple classical calculation turns out to give the right result for the frequency change. a) To determine the frequency of the initial 14 keV photon, convert the energy into joules and, using the relation between energy and frequency of photon $E=h f$, calculate the frequency. b) If we associate a fictitious mass $m$ to the gamma ray photon, given by $m=\frac{E}{c^{2}}$ then the energy change of the photon as it falls in the earth's field through a distance $d$ is given by the familiar potential energy change in a gravitational field, $\Delta E=m g \Delta d$. Express this as a change of frequency of the gamma ray photon $\Delta f$. c) If the distance the photon falls is 22.5 m, calculate both the change in frequency and the fractional change in frequency $\frac{\Delta f}{f}$ of the gamma ray photon. d) This small frequency change is detected by using the Doppler effect in which a moving source emits a wave whose frequency is modified by its motion. The fractional change of frequency emitted is given by the ratio $v / c$ where $v$ is the speed of the source required. Calculate $v$.

speed of light, $c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$ Planck's constant, $h=6.6 \times 10^{-34} \mathrm{Js}$ $e=1.6 \times 10^{-19} \mathrm{C}$

Show worked solution

This problem involves testing General Relativity by measuring gravitational redshift.

The physics concept:

According to Einstein's General Theory of Relativity, light gains energy when falling in a gravitational field (like any object would). For light:
- Energy is related to frequency: $E = hf$ (Planck's relation)
- Higher energy = higher frequency
- A photon "falling" should increase in frequency
Given data:
- Photon energy: $E = 14$ keV $= 14 \times 10^3$ eV
- Falling distance: $d = 22.5$ m
- Speed of light: $c = 3.0 \times 10^8$ m/s
- Planck's constant: $h = 6.6 \times 10^{-34}$ J$\cdots$
- Elementary charge: $e = 1.6 \times 10^{-19}$ C
a) Calculating the initial frequency:

Convert energy to joules: $$E = 14 \times 10^3 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV}$$

$$E = 2.24 \times 10^{-15} \text{ J}$$

Calculate frequency using $E = hf$: $$f = \frac{E}{h} = \frac{2.24 \times 10^{-15}}{6.6 \times 10^{-34}}$$

$$f = 3.4 \times 10^{18} \text{ Hz}$$

This is a gamma ray in the X-ray/gamma region of the electromagnetic spectrum.

b) Deriving the frequency change formula: Fictitious mass of photon: Using $E = mc^2$, the effective mass is: $$m = \frac{E}{c^2} = \frac{hf}{c^2}$$ Energy change as photon falls distance $d$: $$\Delta E = mgd = \frac{hf}{c^2} \cdot g \cdot d$$ Convert to frequency change: $$\Delta E = h\Delta f$$ Therefore: $$h\Delta f = \frac{hfgd}{c^2}$$

$$\Delta f = \frac{fgd}{c^2}$$

This formula shows the frequency change depends on initial frequency, gravity, height, and inversely on $c^2$.

c) Calculating the frequency change: Substitute values: $$\Delta f = \frac{(3.4 \times 10^{18})(9.8)(22.5)}{(3.0 \times 10^8)^2}$$

$$\Delta f = \frac{7.5 \times 10^{20}}{9.0 \times 10^{16}}$$

$$\Delta f = 8.3 \times 10^3 \text{ Hz} = 8.3 \text{ kHz}$$

Fractional change: $$\frac{\Delta f}{f} = \frac{8.3 \times 10^3}{3.4 \times 10^{18}}$$

$$\frac{\Delta f}{f} = 2.5 \times 10^{-15}$$

This is an incredibly small change - only about 2.5 parts in a quadrillion!

d) Calculating the required velocity:

The Doppler effect tells us that: $$\frac{\Delta f}{f} = \frac{v}{c}$$

Solve for velocity: $$v = c \times \frac{\Delta f}{f}$$

$$v = (3.0 \times 10^8) \times (2.5 \times 10^{-15})$$

$$v = 7.4 \times 10^{-7} \text{ m/s}$$

$$v = 0.74 \text{ micrometers per second}$$

Physical interpretation:

To detect this tiny frequency shift due to gravity, Pound and Rebka needed to measure the Doppler shift equivalent to moving at less than 1 micrometer per second - an incredibly sensitive measurement! This experiment was a brilliant confirmation of General Relativity's prediction that gravity affects light.

2015-10II · Long answerd4Modern Physics · radiation pressure and photon momentum

The Sun emits light (and other parts of the electromagnetic spectrum). The light can be described in terms of particles called photons. The energy of a single photon $E_{p h}$ is given by $E_{p h}=h f$, where $h$ is Planck's constant and $f$ is the frequency of the light. The photon also has a momentum, somewhat like the particles in a gas, and will produce a force $F$ on a reflecting surface, given by $F=2 E_{p h} / c$ (the factor "2" is because the photons arrive and get reflected back).

A spacecraft can be driven by a 'solar sail', of unknown dimensions, which consists of a large sheet of reflective material kept facing the Sun. When a photon from the Sun hits the sail it is reflected off and, as a result, the sail experiences a force. Consider one square metre of area. If $n$ photons (per square metre) are reflected each second, then the average force will be given by

$$F_{a v}=\frac{2 n E_{p h}}{c} \text { per square metre. }$$

As photons spread out radially from the Sun, the intensity of photons (the number crossing each square metre of a sphere surrounding the Sun every second) follows an inverse square law, i.e. $n$ is proportional to $\frac{1}{r^{2}}$, where $r$ is the distance measured from the centre of the Sun.

A solar sail fixed to a spacecraft, with the full area of the sail facing the Sun, causes an acceleration of $1.2 \mathrm{mms}^{-2}$ when the spacecraft is far from the Sun, crossing the orbit of Jupiter. With this low mass satellite you can ignore any other effects.

Assume that the photons correspond to a wavelength of $549 \times 10^{-9} \mathrm{~m}$. The total number of photons reaching the Earth from the Sun, $n_{\mathrm{E}}=3.6 \times 10^{21} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$.

Data: $\lambda_{\text {light }}=549 \times 10^{-9} \mathrm{~m}$, Distance Sun - Earth $=150 \times 10^{6} \mathrm{~km}$, Distance Sun - Jupiter $=780 \times 10^{6} \mathrm{~km}$, Mass of spacecraft $=150 \mathrm{~kg}$

a) Calculate the area of the sail, $\boldsymbol{A}$.

b) If the mass of the spacecraft was about five times greater, the acceleration drops to almost zero. This is true at Jupiter's orbital distance from the Sun or the Earth's orbital distance, or in fact at any distance from the Sun. Why is this?

Show worked solution

This problem involves solar sails, radiation pressure, and space propulsion.

Understanding the physics:

Light carries momentum and exerts pressure on surfaces. When reflected, the momentum change is twice the incident value.

Photon momentum and force: Momentum of one photon: $$p_{ph} = \frac{E}{c} = \frac{hf}{c}$$ Force from reflection: $$F = \frac{\Delta p}{\Delta t} = \frac{2p_{ph}}{\Delta t}$$

For $n$ photons per second reflecting: $$F = \frac{2nE_{ph}}{c} = \frac{2nhf}{c}$$

Inverse square law:

Photon intensity follows: $$n \propto \frac{1}{r^2}$$

At different distances from Sun: $$\frac{n_J}{n_E} = \frac{R_E^2}{R_J^2}$$

a) Calculating sail area: Given:
- Spacecraft mass: $m = 150$ kg
- Acceleration at Jupiter: $a = 1.2$ mm/s$^2$ $= 1.2 \times 10^{-3}$ m/s$^2$
- Wavelength: $\lambda = 549 \times 10^{-9}$ m
- Photons at Earth: $n_E = 3.6 \times 10^{21}$ m$^{-2}$s$^{-1}$
- Distances: $R_E = 150 \times 10^6$ km, $R_J = 780 \times 10^6$ km
Energy per photon: $$E_{ph} = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{549 \times 10^{-9}}$$

$$E_{ph} = 3.6 \times 10^{-19} \text{ J}$$

Photon flux at Jupiter: $$n_J = n_E \times \frac{R_E^2}{R_J^2} = 3.6 \times 10^{21} \times \frac{150^2}{780^2}$$

$$n_J = 1.33 \times 10^{20} \text{ photons/m}^2/\text{s}$$

Force equation: $$F = ma = \frac{2n_J E_{ph} A}{c}$$

$$A = \frac{mac}{2n_J E_{ph}} = \frac{150 \times 1.2 \times 10^{-3} \times 3 \times 10^8}{2 \times 1.33 \times 10^{20} \times 3.6 \times 10^{-19}}$$

$$A = 5.6 \times 10^5 \text{ m}^2$$

This is approximately $750 \times 750$ m!

b) Why higher mass reduces acceleration:

The solar sail provides constant thrust regardless of spacecraft mass.

Sun's gravitational pull: $$F_{grav} = \frac{GMm}{r^2}$$

For 5$\times$ mass (750 kg):
- Gravitational force increases by $5\times$
- Solar sail thrust remains the same
- At some point: $F_{grav} \approx F_{sail}$
The sail approaches zero net acceleration when gravitational attraction equals radiation pressure!

For this specific design, with 5$\times$ the mass, the acceleration drops nearly to zero because:
- Increased mass means more gravitational attraction
- Constant sail thrust can't overcome stronger gravity
- System approaches equilibrium

2017-13II · Long answerd5Modern Physics · gravitational redshift — photon energy and frequency shift

The announcement in 2016 about the discovery of gravitational waves has demonstrated the effects that gravity has in more subtle ways than just dropping a weight on our toe. The Einstein Tower gedanken (thought) experiment illustrates that electromagnetic waves are affected by gravity. It uses the equation for the equivalence of mass and energy, $E=m c^{2}$ in which $c$ is the speed of light, $m$ is the mass of an object, and $E$ is the amount of energy into which it can in principle be converted.
a) If a small mass $m$ is dropped from a height $h$, it gains kinetic energy as it falls. On reaching the ground, it could all (in principle) be converted into electromagnetic waves in the form of photons (packets of energy), each of energy $E=h f$, and reflected back up. The photons could then be converted back into a mass. Explain using the concept of energy conservation, why this process indicates that photons must be affected by gravity.
b) A photon of frequency of $4.2 \times 10^{14} \mathrm{~Hz}$ is emitted towards the ground from a satellite orbiting the Earth at a height of 450 km . Assume that the acceleration due to gravity, g is constant and the motion of the satellite may be ignored in this calculation. By identifying the energy of the photon as it leaves the satellite with a fictitious mass $m_{\gamma}$, calculate the change in frequency, $\Delta f$, of the photon when it reaches the ground.
c) When the photon travels the 450 km to Earth, how many wavelengths of light would this be if the photon was not affected by gravity?
d) Since the photon is affected (slightly) by gravity, comment on how the wavelength changes as the photon falls.
e) What is the change in the number of complete wavelengths along the 450 km path due to the effect of gravity on the photon?

Show worked solution

a) Mass falls and gains kinetic energy. Converted to photons, reflected back, and converted to greater mass. If mass was dropped again, there would be more energy to convert, implying energy increases indefinitely. This violates energy conservation unless photons are affected by gravity (gaining energy falling down, losing energy going up).

b) $hf_{ground} = hf_{sat}\left(1 + \frac{gH}{c^{2}}\right)$

$\Delta f = f_{ground} - f_{sat} = \frac{f_{sat}gH}{c^{2}} = 4.2 \times 10^{14} \times 9.81 \times 450000 / (9 \times 10^{16}) = 20600$ Hz $= 21$ kHz

c) Number of wavelengths: $N = \frac{450000}{c/f} = \frac{450000}{3 \times 10^{8} / 4.2 \times 10^{14}} = 6.1 \times 10^{11}$

d) Since $\Delta f \propto H$, wavelength will uniformly reduce as photon approaches ground.

e) Extra waves: $\Delta N = \Delta t \cdot \frac{\Delta f}{2} = \frac{450000}{3 \times 10^{8}} \times \frac{20600}{2} \approx 15$

2020-10II · Long answerd4Modern Physics · photon energy and electron acceleration voltage

To study the structure of crystals, X-rays with wavelength approximately equal to the size of an atom are used. For sodium: density $\rho_{m} = 0.971 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, molar mass $M = 23.0 \mathrm{~g} \mathrm{~mol}^{-1}$.
a) Assuming a simple cubic arrangement of sodium atoms in the crystalline structure, estimate the diameter of a sodium atom.
b) What would be the energy of a photon with this wavelength?
c) Hence determine the accelerating voltage required to produce such photons.

Show worked solution

This problem involves X-ray crystallography.

Given data for sodium:
- Density: $\rho = 0.971 \times 10^3 kg/m^3$
- Molar mass: $M = 23.0$ g/mol = $0.023 kg/mol$
a) Estimating atomic diameter: Volume per atom: $$V_{atom} = \frac{M}{\rho N_A} = \frac{0.023}{971 \times 6.02 \times 10^{23}}$$

$$V_{atom} = 3.94 \times 10^{-29} \text{ m}^3$$

Assuming cubic packing: $$d^3 = V_{atom}$$

$$d = (3.94 \times 10^{-29})^{1/3} = 3.40 \times 10^{-10} \text{ m}$$

$$d \approx 0.34 \text{ nm}$$

b) Photon energy: $$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{0.34 \times 10^{-9}}$$

$$E = 5.85 \times 10^{-16} \text{ J}$$

$$E = \frac{5.85 \times 10^{-16}}{1.6 \times 10^{-19}} = 3655 \text{ eV}$$

c) Accelerating voltage: $$eV = E$$

$$V = \frac{3655}{1} = 3655 \text{ V}$$

Answer: D (3655 V)