Physics (Theory)

1. When the junction is formed, the n-side has many free electrons and the p-side has many holes, so there is a concentration difference across the junction.
2. Electrons diffuse from the n-side to the p-side and holes diffuse from the p-side to the n-side (diffusion current).
3. An electron leaving the n-side leaves behind an immobile positive donor ion; a hole leaving the p-side leaves behind an immobile negative acceptor ion.
4. So a thin layer of fixed positive ions forms on the n-side and of fixed negative ions on the p-side of the junction. This layer has no free charge carriers – it is called the depletion region (width ~ 1 μm).
5. The ions create an electric field (and a barrier potential) directed from n to p, which opposes further diffusion and drives a drift current. Equilibrium is reached when drift current = diffusion current.
Tip: a labelled sketch showing + ions on the n-side and – ions on the p-side earns marks too.

(i)(a) Impact parameter b: the perpendicular distance of the initial velocity vector (line of motion) of the α-particle from the centre of the nucleus.
(b) Distance of closest approach \(r_0\): in a head-on approach, the minimum distance from the centre of the nucleus that the α-particle reaches, where all its kinetic energy K has become electrostatic potential energy: \(K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(Ze)(2e)}{r_0}\Rightarrow r_0 = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}\).
(ii) Smaller impact parameter → larger scattering (b ∝ cot(θ/2)).
(I) θ = 0° (undeviated): b is very large – the particle passes far from the nucleus (maximum, of the order of atomic size).
(II) θ = 180° (bounces straight back): b = 0, i.e. a head-on collision.

Einstein's equation: \(K_{max} = h\nu - h\nu_0\).
Case 1: \(K = h\nu_1 - h\nu_0\) …(1)
Case 2: \(2K = h\nu_2 - h\nu_0\) …(2)
Multiply (1) by 2 and equate with (2): \(2h\nu_1 - 2h\nu_0 = h\nu_2 - h\nu_0\)
\(\Rightarrow h\nu_0 = 2h\nu_1 - h\nu_2\Rightarrow \nu_0 = 2\nu_1 - \nu_2\).
Check: since 2K > K, \(\nu_2 > \nu_1\), and \(\nu_0 < \nu_1\) requires \(\nu_2 < 2\nu_1\) – consistent.

1. Generation: photons with energy hν greater than the band gap E_g are absorbed close to the p-n junction and create electron–hole pairs.
2. Separation: the electric field of the depletion region pushes the electrons towards the n-side and the holes towards the p-side.
3. Collection: electrons reaching the n-side are gathered by the front contact and holes reaching the p-side by the back contact.
As a result the p-side becomes positive and the n-side negative, so a photo-voltage (emf) appears across the cell and a current flows through an external load.
Marks: one point for each process plus the conclusion about the polarity of the cell.

(i) Sketch: x-axis as direction of travel; E drawn as a sine wave in the x-y plane (along y) and B as a sine wave in the x-z plane (along z), both in phase and perpendicular to each other and to x.
Expressions: \(E_y = E_0\sin(kx - \omega t)\), \(B_z = B_0\sin(kx - \omega t)\), with \(k = 2\pi/\lambda\), \(\omega = 2\pi\nu\) and \(E_0/B_0 = c\).
(ii) Any three:
• They are transverse – E and B are perpendicular to each other and to the direction of propagation.
• In vacuum all travel with speed \(c = 1/\sqrt{\mu_0\varepsilon_0} = 3\times10^8\) m/s.
• They carry energy and momentum (and exert radiation pressure).
• They need no material medium.
• E and B oscillate in phase.

(i) Radioactive decay of nuclei → gamma rays. Use: radiotherapy to destroy cancer cells / sterilising medical equipment.
(ii) Welding arcs → ultraviolet rays (they also give visible and infrared light). Use: killing germs in water purifiers; LASIK eye surgery. (That is why welders wear special goggles.)
(iii) Hot bodies → infrared rays (heat radiation). Use: night-vision cameras, TV remote controls, physiotherapy heating.

(a) Conditions for TIR:
1. Light must travel from an optically denser medium towards a rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.
(b)(i) For near-normal viewing, apparent depth = real depth / μ. So apparent depth ∝ real depth: the graph is a straight line through the origin (real depth on x-axis from 0 to H, apparent depth on y-axis), with slope less than 1.
(ii) Slope = apparent depth / real depth = \(1/\mu\), the reciprocal of the refractive index of the liquid. So μ = 1/slope.
Common mistake: plotting slope = μ – the slope is less than 1, so it is 1/μ.

Ray diagram: parallel rays from a distant object, inclined at angle α to the axis, fall on the objective (focal length \(f_o\)) and form a real, inverted image of height h at its focal point, which is also the focus of the eyepiece (\(f_e\)). Rays leave the eyepiece parallel, making angle β with the axis, so the final image is at infinity. Label objective, eyepiece, \(f_o\), \(f_e\), intermediate image, α, β, arrows on rays.
Magnifying power \(m = \beta/\alpha\) (angle subtended by final image / angle subtended by object).
For small angles: \(\alpha \approx \tan\alpha = h/f_o\) and \(\beta \approx \tan\beta = h/f_e\).
\(\Rightarrow m = \dfrac{h/f_e}{h/f_o} = \dfrac{f_o}{f_e}\) (with sign, \(-f_o/f_e\), as the image is inverted).
In normal adjustment the length of the telescope is \(f_o + f_e\).
Tip: half a mark is lost if arrows or labels are missing.

(i) Intensity graph: a tall, broad central maximum at θ = 0, with zeros at \(\theta = \pm\lambda/a, \pm2\lambda/a,\dots\); between them much weaker secondary maxima near \(\pm3\lambda/2a, \pm5\lambda/2a\), whose intensity falls off rapidly.
(ii) Divide the slit into three equal parts. When path difference between the two edges is \(a\sin\theta = 3\lambda/2\), the first two thirds cancel each other (path difference λ/2 between corresponding points), and the last third adds up → a (weak) maximum.
General condition for secondary maxima: \(a\sin\theta = (n + \tfrac12)\lambda\); first maximum, n = 1: \(a\theta = 3\lambda/2\) (small θ).
\(\theta = \dfrac{3\lambda}{2a}\); distance on screen \(x = D\theta = \dfrac{3\lambda D}{2a}\).

Lens maker's formula: \(\dfrac1f = \left(\dfrac{\mu_g}{\mu_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). A converging (biconvex) lens: \(R_1 = +10\) cm, \(R_2 = -20\) cm, so \(\dfrac1{R_1}-\dfrac1{R_2} = \dfrac1{10}+\dfrac1{20} = \dfrac3{20}\) cm⁻¹.
(a) In air: \(\dfrac1f = (1.5-1)\times\dfrac3{20} = \dfrac3{40}\Rightarrow f = \dfrac{40}3 \approx 13.3\) cm.
(b) In liquid: \(\dfrac{\mu_g}{\mu_l} = \dfrac{1.5}{1.25} = 1.2\); \(\dfrac1{f_l} = (1.2-1)\times\dfrac3{20} = \dfrac{3}{100}\Rightarrow f_l = \dfrac{100}3 \approx 33.3\) cm.
The lens is still converging but weaker, since the liquid's refractive index is closer to that of glass.

First excited state means n = 2.
(a) \(r_n = n^2 r_0 = 4\times0.53 = 2.12\) Å.
(b) Bohr's quantisation: \(L = \dfrac{nh}{2\pi} = \dfrac{2\times6.63\times10^{-34}}{2\times3.14} = \dfrac{6.63\times10^{-34}}{3.14}\approx 2.11\times10^{-34}\) J s (kg m² s⁻¹).
(c) In a Bohr orbit, K = −E and U = 2E.
K = +3.4 eV; U = 2 × (−3.4) = −6.8 eV. (Check: K + U = −3.4 eV.)
Common mistake: taking n = 1 for 'first excited' – the ground state is n = 1.

(a) Graph of U vs r: for r greater than about 0.8 fm the PE is negative and has a minimum near r₀ ≈ 0.8 fm (strongly attractive force); beyond ~2–3 fm it rises quickly to zero (short range). For r < 0.8 fm the PE rises steeply and becomes positive (strong repulsion).
(b) Q = [m(Fe) − 2 m(Al)] c².
2 m(Al) = 2 × 27.98191 = 55.96382 u.
Δm = 55.93494 − 55.96382 = −0.02888 u.
Q = −0.02888 × 931.5 MeV ≈ −26.9 MeV.
Q is negative: the products have more mass than the parent, so energy would have to be supplied. The fission is not energetically possible (Fe-56 is near the peak of the binding-energy curve and is very stable).
Note: some printed keys show the mass difference as −0.02442 u; the correct arithmetic gives −0.02888 u – the conclusion is the same.

Use \(m_\alpha = 4m_p\), \(q_\alpha = 2q_p\).
(a) \(\lambda = \dfrac{h}{mv}\): same v ⇒ \(\dfrac{\lambda_\alpha}{\lambda_p} = \dfrac{m_p}{m_\alpha} = \dfrac14\).
(b) \(\lambda = \dfrac{h}{\sqrt{2mK}}\): same K ⇒ \(\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_p}{m_\alpha}} = \dfrac12\).
(c) Through potential V, K = qV, so \(\lambda = \dfrac{h}{\sqrt{2mqV}}\) ⇒ \(\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\dfrac{1}{4\times2}} = \dfrac{1}{2\sqrt2}\).
Common mistake: forgetting that the α-particle has double charge in part (c).

Circuit: AC input to the primary of a centre-tapped transformer. Secondary ends A and B connect to the p-sides of diodes D₁ and D₂; their n-sides are joined and go to one end of the load resistor R_L; the other end of R_L goes to the centre tap.
Working:
• Positive half-cycle: A is positive and B negative relative to the centre tap. D₁ is forward biased and conducts; D₂ is reverse biased. Current flows through R_L in a fixed direction.
• Negative half-cycle: A is negative and B positive. Now D₂ conducts and D₁ is off. Current through R_L flows in the same direction as before.
So output is obtained during both halves of the input cycle.
Waveforms: input – a full sine wave; output – a series of positive half-sine humps with no gaps (D₁ and D₂ humps alternate), i.e. pulsating unidirectional voltage, frequency twice the input.