June 2019
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Section A
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Answer all questions in this section.
- 1.1) Two isotopes of iodine are [math]_{53}^{125}I[/math] and [math]_{53}^{131}I[/math]
Determine, for these two isotopes, the difference between the constituents of the nuclei. - Explanation:
Iodine – 125 (I-125):
Protons: [math]53[/math] (since it’s iodine)
Neutrons: [math]125 – 53 = 72[/math] - Iodine – 131 (I – 131):
Protons: [math]53[/math]
Neutrons: [math]72 + 6 = 78[/math] (6 more neutrons than [math]I – 125[/math]) - In summary, Iodine-131 has 6 more neutrons than Iodine-125, with both isotopes having 53 protons but differing neutron counts ([math]72[/math] neutrons for [math]I – 125[/math] and [math]78[/math] neutrons for [math]I- 135[/math]).
- 1,2) A [math]_{53}^{131}I[/math] nuclide undergoes beta ([math]\beta^-[/math]) decay to form a xenon nuclide.
- State the nucleon number of the xenon nuclide.
- Explanation:
- In beta ([math]\beta^-[/math]) decay, a neutron is transformed into a proton, and an electron ([math]\beta^-[/math] particle) and an antineutrino are emitted. The atomic number increases by one, while the mass number remains the same.
- Given that Iodine-131 ([math]^{131}I[/math]) undergoes [math]\beta^-[/math] decay, the resulting nuclide is Xenon. Since the atomic number (proton number) increases by one in the process, and the mass number (nucleon number) remains the same, the nucleon number of the Xenon nuclide will also be [math]131[/math].
- 1.3) A [math]_{53}^{125}I[/math] nuclide decays by electron capture to form a tellurium nuclide.
- State two differences between the constituents of the iodine nucleus and the tellurium nucleus it decays into.
- Explanation:
I. Proton content: - Iodine-125 ([math]^{125}I[/math]) has [math]53[/math] protons in its nucleus.
- Tellurium, formed after electron capture, has one fewer proton, so it has [math]53 – 1 = 52[/math] protons.
- The reduction in proton number is a result of electron capture, where a proton in the nucleus captures an electron, transforming into a neutron.
- II. Neutron Content:
- Iodine-125 ([math]^{125}I[/math]) has [math]125 – 53 = 72[/math] neutrons in its nucleus.
- Tellurium, formed after electron capture, has one more neutron, so it has [math]72 + 1 = 73[/math] neutrons.
- The increase in neutron number is a result of the conversion of a proton into a neutron during electron capture.
- In summary: electron capture in iodine-125 results in the formation of tellurium with two key differences:
- – One fewer proton in the tellurium nucleus compared to iodine-125 due to the conversion of a proton into a neutron.
- – One more neutron in the tellurium nucleus compared to iodine-125 due to the same conversion process.
- 1.4) Internal conversion is a process in which a nucleus in an excited state can release its excess energy. In internal conversion all of the excess energy is transferred from the nucleus to an orbital electron through the electromagnetic force. This orbital electron is ejected from the atom.
- The tellurium nucleus formed in question 01.3 is in an excited state and can undergo internal conversion.
Discuss three differences between internal conversion and beta ([math]\beta^-[/math]) decay. - Explanation:
I. Particle released: - – Internal conversion:
- Only an electron is released. The excess energy from the excited nucleus is transferred directly to an orbital electron, which is then ejected from the atom.
- – Beta ([math]\beta^-[/math]) decay:
- In beta decay, both an electron (beta particle) and an antineutrino are released. The antineutrino is a neutral particle that escapes from the nucleus, and the electron is emitted.
- II. Energies / momenta of released particles:
- – Internal conversion:
- Electrons released in internal conversion have similar and discrete energies and momenta. The energy transfer is direct, resulting in a specific energy level for the ejected electron.
- – Beta ([math]\beta^-[/math]) decay:
- Electrons released in beta decay have a range of energies and momenta. The energy distribution is broader compared to internal conversion, reflecting the continuum of possible energy states in the decay process.
- III. Effect on Nucleus / Element:
- – Internal conversion:
- There is no change in the constituents of the nucleus during internal conversion. The element does not change; it remains the same.
- – Beta ([math]\beta^-[/math]) decay:
- In beta decay, a neutron is converted into a proton. This results in a change in the element because the number of protons in the nucleus increases by one. The identity of the element changes, leading to a different atomic number.
- In summary, these differences highlight key aspects of the two processes, including the particles involved, the energy characteristics of the emitted particles, and the impact on the nucleus and element.
- 2. Some cars are fitted with a water sensor designed to switch on windscreen wipers automatically when it rains. Figure 1 shows a simplified diagram of the sensor.
- Explanation:
- A light ray travels from the light-emitting diode (LED) through the first prism and into the windscreen. The ray reflects off the surfaces of the windscreen at A, B and C and then passes through the second prism into the detector.
- 2.1) Suggest how the design ensures that there is no deviation of the ray as it enters the first prism.
- Explanation:
Enters prism along normal:
– When the light ray enters the first prism along the normal, it means that the incident angle is zero degrees concerning the normal. According to Snell’s Law, the formula for refraction is - [math] n_1 \sin(\theta_1) = n_2 \sin(\theta_2) [/math]
- Where [math] n_1 [/math] and [math] n_2 [/math] are the refractive indices of the two media, and [math] \theta_1 [/math] and [math] \theta_2 [/math] are the angles of incidence and refraction, respectively.
- – Since the incident angle [math] \theta_1 = 0^\circ [/math] (along the normal), [math] \sin(0^\circ) = 0 [/math]. Therefore, there is no change in direction of the light ray as it enters the first prism.
- 2.2) Suggest two features of the design that ensure that there is no deviation of the ray as it leaves the first prism and enters the windscreen glass.
- Explanation:
- If the first prism and the windscreen glass have the same refractive index and are in direct contact (touching with no gap between them), this design configuration ensures that there is no deviation of the ray as it leaves the first prism and enters the windscreen glass.
- When two optical materials with the same refractive index are in contact, the light ray will not undergo any bending or deviation at the interface. This is because Snell’s Law, which describes how light is refracted at an interface between two media, involves the ratio of the refractive indices of the two media. If the refractive indices are the same, the ratio becomes 1, and according to Snell’s Law:
- [math] n_1 \sin(\theta_1) = n_2 \sin(\theta_2) [/math]
- If [math] n_1 = n_2 [/math], the equation simplifies to:
- [math] \sin(\theta_1) = \sin(\theta_2) [/math]
- Since the refractive indices are the same, there is no change in direction ([math] \theta_1 = \theta_2 [/math])when the light ray passes from the first prism to the windscreen glass. This ensures the ray continues without deviation.
- 2.3) The refractive index of the windscreen glass is 1.52. Explain why the ray follows the path shown inside the windscreen glass in Figure 1. Support your answer with a suitable calculation.
- Explanation:
- The path of the ray inside the windscreen glass, as shown in Figure 1, can be explained by the phenomenon of total internal reflection (TIR). Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence exceeds the critical angle.
- In this case, the light ray is traveling from the windscreen glass (higher refractive index) to the air outside (lower refractive index). The critical angle (C) can be calculated using Snell’s Law:
- [math] Critical\ angle\ (C) = \arcsin\left(\frac{n_{lower}}{n_{higher}}\right) [/math]
- Where:
- – n_lower is the refractive index of the lower refractive index medium (air in this case)
- – n_higher is the refractive index of the higher refractive index medium (windscreen glass).
- Substituting the given values:
- [math] (C) = \arcsin\left(\frac{n_{lower}}{n_{higher}}\right) [/math]
[math] (C) = \arcsin\left(\frac{1}{1.52}\right) [/math]
[math] (C) = \arcsin(0.6579) [/math]
[math] C = 41.1^{\circ} [/math] - The calculation confirms that the critical angle (C) is approximately 41.1. Since the incident angle inside the windscreen glass is greater than the critical angle (45° > 41.1°), total internal reflection occurs at each boundary between the glass and air. This means that the light ray reflects completely within the glass, following the path shown in Figure 1, rather than being transmitted into the air. Total internal reflection is a crucial phenomenon in various optical applications, such as fiber optics and prismatic devices.
- 2.4) When it starts to rain, water droplets form on the outside of the windscreen as shown in Figure 2
- – The refractive index of water is 1.33
– Explain why the presence of water at A causes the intensity of the light at the detector to decrease.
– Support your answer with a suitable calculation. - Explanation:
- The decrease in the intensity of light at the detector when water is present at point A on the windscreen can be explained by the change in the optical conditions at the glass-water interface. The key factor is the critical angle, which determines whether total internal reflection occurs or if some light is refracted into the water.
- I. Calculation of critical angle at glass – water boundary:
– Use Snell’s Law to calculated critical angle (C) at the glass – water boundary: - [math] (C) = \arcsin\left(\frac{n_{lower}}{n_{higher}}\right) [/math]
[math] (C) = \arcsin\left(\frac{1.52}{1.33}\right) [/math]
[math] (C) = \arcsin(0.875) [/math]
[math] (C) = 61.0^{\circ} [/math] - II. Analysis of critical angle:
– The calculated critical angle is [math] 61.0^{\circ} [/math]
– This means that when light travels from the windscreen glass to water and the incident angle exceeds [math] 61.0^{\circ} [/math], total internal reflection no longer takes place. - III. Calculation of angle of refraction in water:
– If the incident angle inside the glass is greater than the critical angle, some light will be refracted into the water. - [math] Angle\ of\ refraction = 90^{\circ} – C [/math]
[math] Angle\ of\ refraction = 90^{\circ} – 61.0^{\circ} [/math]
[math] Angle\ of\ refraction = 28.9^{\circ} [/math] - IV. Explanation for intensity decrease:
– As the incident angle is greater than the critical angle, some light refracts into the water.
– Total internal reflection no longer occurs, so less light reflects within the windscreen glass.
– Some light escapes/refracts into the water droplets on the outside of the windscreen.
Therefore, less light stays within the windscreen, resulting in a decrease in the intensity of light detected at the sensor. - 2.5) The refractive index of the windscreen glass can vary by a few per cent across the thickness of the glass. Discuss how this variation may affect the path of the ray through the windscreen glass.
- Explanation:
- I. Non – straight path:
- As the refractive index varies, the trajectory of the light ray through the glass may not be a straight line.
- Variations in n can lead to bending and curving of the ray, impacting its overall path.
- Intensity changes:
- – Non-uniform paths may result in some portions of the light being deviated more than others.
- – This deviation could lead to scattering or spreading of the light within the glass.
- – Consequently, the intensity of the light reaching the detector may vary spatially, affecting the overall reading.
- In conclusion, even a small percentage variation in the refractive index of the windscreen glass can lead to non-straight paths for light rays and consequential changes in intensity at the detector. The significance of these effects depends on the specific application and the tolerances of the detection system in use.
- 2.6) A different design has the LED and the detector further apart. The ray undergoes more reflections inside the windscreen glass before reaching the detector. Discuss two ways in which this design affects sensitivity of the sensor.
- Explanation:
- More Sensitive:
- I. Increased Encounter with water droplets:
- The design with the LED and detector further apart results in a longer path for the light ray inside the windscreen.
- A longer path increases the chances of the light encountering water droplets, contributing to a more sensitive sensor.
- This design is more likely to detect water droplets as it offers more opportunities for the light to refract out in the presence of rain.
- II. Surface imperfections and scratches:
- With the light undergoing more reflections inside the windscreen, any imperfections on the glass surface, such as scratches, become more significant.
- These surface imperfections can act as additional points for light to scatter or refract increasing the likelihood of encountering and detecting water droplets.
- The sensor becomes more sensitive as it leverages these imperfections to enhance its ability to detect rain.
- Less Sensitive:
- I. Surface imperfections and scratches:
- While surface imperfections enhance sensitivity to water droplets, they can also introduce noise in the absence of rain.
- Scratches and imperfections may scatter or absorb light even when there are no water droplets, leading to false detections or a reduced baseline sensitivity.
- The sensor may become less reliable in distinguishing between rain-related and non-rain-related signals.
- II. Intensity reduction without droplets:
- The increased path length inside the windscreen can lead to greater absorption of light by the glass material itself.
- In the absence of water droplets, the longer path results in a more significant reduction in light intensity, potentially lowering the sensitivity of the sensor to subtle rain events.
- The design may be less sensitive in scenarios where the changes in light intensity are not solely attributed to the presence of water droplets
- 3) Figure 3 shows an arrangement to investigate diffraction. White light is incident on a single slit. After leaving the slit, the diffracted light passes through a green filter to reach the screen.
- 3.1) Describe the pattern produced on the screen.
- Explanation:
- The pattern produced on the screen in this diffraction setup would exhibit a central maximum with lower intensity maxima on either side. The central maximum is characterized by being twice as wide or wider than the other maxima.
Here’s a breakdown of the description: - I. Central maximum:
– The central maximum is the primary, central region of the diffraction pattern.
– It has the highest intensity or brightness.
– The width of the central maximum is emphasized as being twice as wide or wider compared to the other maxima. - II. Lower Intensity maxima (Either side):
– On either side of the central maximum, there are additional maxima.
– These side maxima have lower intensity or brightness compared to the central maximum.
– The diffraction pattern typically exhibits a series of alternating maxima and minima. - In summary, the diffraction pattern on the screen is characterized by a central maximum that is broader than the other maxima, and on either side of the central maximum, there are lower intensity maxima.
- This pattern is a result of the diffraction of white light through a single slit and the subsequent filtering with a green filter.
- 3.2) The green filter is replaced with a red filter. Describe the change in the pattern produced on the screen.
- Explanation:
- When the green filter is replaced with a red filter in the diffraction setup, the pattern produced on the screen undergoes a change. The primary changes in the diffraction pattern are:
- I. Wider central Maximum:
– The central maximum becomes wider compared to the previous setup with the green filter.
– The change in filter color affects the diffraction of light, resulting in a broader central maximum on the screen. - II. Subsequent Maxima further Apart:
– The maxima and minima in the diffraction pattern, which occur on either side of the central maximum, become further apart.
– The alteration in the color of the filter influences the diffraction characteristics of the light, causing subsequent maxima to be spaced more widely. - In summary, the replacement of the green filter with a red filter leads to a diffraction pattern with a wider central maximum and subsequent maximum that are further apart on the screen. The specific wavelengths associated with red light compared to green light influence the diffraction angles and, consequently, the observed pattern.
- 3.3) A diffraction grating is placed between the red filter and the screen. The diffraction grating has 500 lines per millimetre. Light is incident normally on the grating. Figure 4 shows the arrangement.
- The wavelength of the red light is 650 nm.
Calculate the angle θ between a first-order maximum and the central maximum. - Explanation:
- The angle (θ) between a first-order maximum and the central maximum in a diffraction grating can be calculated using the grating equation.
- [math] \sin\theta = \frac{n\lambda}{d} [/math]
- Where:
– θ is the angle between the diffracted beam and the central maximum,
– n is the order of the maximum (in this case, the first-order maximum, so n=1)
– λ is the wavelength of light,
– d is the grating spacing (distance between adjacent slits or lines). - Given:
- [math] \lambda = 650\ \text{nm} [/math] (converted to meters)
[math] \lambda = 6.5 \times 10^{-7}\ \text{m} [/math] - [math] d = \frac{1 \times 10^{-3}}{500} = 2 \times 10^{-6}\ \text{m} [/math]
- Now, substitute these values
- [math] \sin\theta = \frac{n\lambda}{d} \\ \sin\theta = \frac{(1)(6.5 \times 10^{-7})}{2 \times 10^{-6}} \\ \sin\theta = 0.325 \\ \theta = \sin^{-1}(0.325) \\ \theta = 18.77^{\circ} [/math]
- – So, the angle between the first-order maximum and the central maximum is approximately [math] 18.77^{\circ} [/math]
- 3.4) In practice, the filter transmits red light with wavelengths in the range 600 nm to 700 nm. Suggest how this affects the appearance of the maxima.
- Explanation:
- The expression n=d/λ and use the provided values:
- [math] n = \frac{d}{\lambda} \\ n = \frac{2.5 \times 10^{-6}}{6 \times 10^{-7}} \\ n = \frac{25}{6} \\ n = 4.17 [/math]
- Now, rounding to the nearest whole number, [math] n \approx 4 [/math]
- The discrepancy in the order of the maximum arises from the difference in the calculated values and the expected value of 4. This discrepancy may be due to the approximation of values or slight variations in the provided information.
- Now, let’s discuss how the order affects the width of the diffraction maxima:
- I. Wider Maxima for higher order (n):
– The expression [math] n = \frac{d}{\lambda} [/math] suggests that as the order of the maximum (n) increases, the value of (n) in the denominator becomes larger.
– As n increases, the diffraction maxima become wider. This is because higher-order maxima correspond to larger angles (θ), leading to a broader diffraction pattern. - II. Proportional relationship:
– The relationship between the order of the maximum and the width of the maxima is proportional. As n increases, the width of the maxima also increases proportionally. - III. Illustration:
– For example, if the order of the maximum is 4, the diffraction maxima will be narrower compared to when the order is 10.
– Higher-order maxima contribute to a broader diffraction pattern, resulting in a more spread-out appearance on the screen. - In summary, the calculated order of the maximum (n=4) aligns with the expected behavior: higher-order maxima result in wider diffraction patterns. The relationship [math] n = \frac{d}{\lambda} [/math] helps illustrate the proportional increase in the width of the maxima as the order increases.
- 4) Figure 5 shows a simplified catapult used to hurl projectiles a long way.
- The counterweight is a wooden box full of stones attached to one end of the beam. The projectile, usually a large rock, is in a sling hanging vertically from the other end of the beam. The weight of the sling is negligible.
- The beam is held horizontal by a rope attached to the frame.
- 4.1)
The catapult is designed so that the weight of the beam and the weight of the empty wooden box have no effect on the tension in the rope.
Suggest how the pivot position achieves this
Explanations:
I. Center of Mass Alignment:
– The pivot is strategically located to coincide with the center of mass of the beam and the wooden box.
II. Balanced Moments:
-Moments due to the weights of the beam and the box are balanced at the pivot point.
-The clockwise moment (box weight) is equal and opposite to the anticlockwise moment (beam weight).
III. Zero Sum of moments:
-The sum of moments is zero at the pivot position, ensuring rotational equilibrium.
-Mathematically, the moments on one side of the pivot are balanced by the moments on the other side.
IV. Consideration of moments:
– The pivot position is chosen with careful consideration of moments, the forces that cause rotation.
– Achieving equilibrium involves balancing the moments to prevent any net rotation.
V. Rotational stability:
– The pivot position contributes to the rotational stability of the catapult, preventing uncontrolled movement.
VI. Negligible effect on rope tensions:
– The design ensures that the weights of the beam and the box have no significant effect on the tension in the rope.
– The pivot position enables a configuration where the forces are balanced, and the system remains stable. - In essence, the pivot position is a critical element in the catapult design, providing a stable and balanced arrangement where the weights of the beam and box do not interfere with the tension in the rope.
- 4.2)
The stones in the counterweight have a total mass of [math]610\ \text{kg}[/math] and the projectile weighs [math]250\ \text{N}[/math].
Calculate the tension in the rope. - Explanation:
- To calculate the tension in the rope, we can use the principle of moments, setting the clockwise moment equal to the anticlockwise moment.
- Given:
- – [math] \text{Total mass of stones in the counterweight } m = 610\ \text{kg}[/math]
– [math] g = 9.81\ \text{m s}^{-2}[/math] - – [math] \text{Distance from the pivot to the center of mass of the stones } r = 1.5\ \text{m}[/math]
– [math] \text{Weight of the projectile, } W = 250\ \text{N}[/math]
– [math] \text{Angle of the sling } \theta = 50^{\circ}[/math]
– [math] \text{Distance from the pivot to the point where the rope is attached to the beam } d = 4.0\ \text{m}[/math] - Now, let’s calculate the clockwise and anticlockwise moments:
- I. Clockwise moment (M_{cW}):
- [math]M_{cW} = m g r[/math]
[math]M_{cW} = (610)(9.81)(1.5)[/math]
[math]M_{cW} = 8976\ \text{N m}[/math] - II. Anticlockwise moment (M_{rcW}):
- [math]M_{rcW} = W d + T \sin\theta \cdot d[/math]
[math]M_{rcW} = (250)(4) + T \sin 50^{\circ} \cdot (4)[/math] - Now, set [math]M_{cW}[/math] equal to [math]M_{rcW}[/math] and solve for [math]T[/math]:
- [math]8976 = (250)(4) + T \sin 50^{\circ} \cdot (4)[/math]
- Solve for [math]T[/math]:
- [math]T \approx 8976 – (250)(4) \over \sin 50^{\circ} \cdot (4)[/math]
- [math]T \approx 2600\ \text{N}[/math]
- Therefore, the tension in the rope (T) is approximately [math]2600\ \text{N}[/math].
- 4.3)
- When the rope is cut, the counterweight rotates clockwise. When the beam is vertical it is prevented from rotating further. The projectile is then released horizontally with a velocity of [math]18\ \text{m s}^{-1}[/math], as shown in Figure 6.
The projectile is released at a height of [math]7.5\ \text{m}[/math] above ground level. 
- Figure 6.
- The range of the catapult is the horizontal distance between the point where the projectile is released to the point where it lands.
- Calculate the range.
- Ignore air resistance
- Explanation:
- Given the vertical motion equation:
- [math]s = ut + \tfrac{1}{2} g t^{2}[/math]
- Where:
– [math]s[/math] is the vertical displacement,
– [math]u[/math] is the initial vertical velocity (which is [math]0[/math] in this case),
– [math]g[/math] is the acceleration due to gravity ([math]9.81\ \text{m s}^{-2}[/math]),
– [math]t[/math] is the time of flight. - You mentioned [math]s = 7.5\ \text{m}[/math], so we can use this to find the time of flight ([math]t[/math]):
- [math]t^{2} = \frac{(7.5)(2)}{9.81}[/math]
[math]t^{2} = 1.53[/math]
[math]t = \sqrt{1.53}[/math]
[math]t = 1.24\ \text{s}[/math] - Now, use this time ([math]t[/math]) to find the horizontal distance [math]d[/math]:
- [math]d = u t[/math]
[math]d = (18)(1.24)[/math]
[math]d = 22.32\ \text{m}[/math] - Therefore, the corrected range of the catapult is approximately [math]22.32\ \text{m}[/math].
- 4.4)
- In another release, the sling is adjusted so that a projectile of the same mass is released just before the wooden beam is vertical. The projectile is not released horizontally:
Discuss the effect this change has on the range of the catapult. - Explanation:
- The change in the release of the projectile, just before the wooden beam is vertical, will have a significant effect on the range of the catapult. Here are the effects and corresponding explanations:
- I. Effect: Range will be greater:
Component of velocity upwards:
– By releasing the projectile just before the wooden beam is vertical, there is now a significant vertical component to the projectile’s velocity.
– The vertical component contributes to the projectile gaining height during its trajectory. - Rock will spend longer in the air:
– The vertical component of velocity enables the projectile to spend more time in the air compared to a purely horizontal release.
– The increased time of flight ([math]t[/math]) contributes to a longer trajectory. - Greater time of flight ([math]t[/math]):
– The projectile spends a longer time in the air due to the upward component of velocity.
– A greater time of flight leads to the rock covering a larger horizontal distance. - Therefore, the range is greater:
– The combined effect of the upward velocity component and the increased time of flight results in a greater horizontal range. - II. Balanced arguments:
- Unchanged range / answer is indeterminate:
– It’s possible that the increased upward velocity and the longer time of flight are offset by other factors, such as the counterweight falling less far before the projectile is released.
– The overall effect on the range may be indeterminate due to the complex interplay of various factors. - In summary, the change in releasing the projectile just before the wooden beam is vertical is likely to increase the range of the catapult due to the introduction of an upward velocity component and a longer time of flight.