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Explain semiconductors, doping, and the conservation of momentum. Calculate the number of electrons that pass through a wire in 4 seconds at 8 A.

Physics

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1. a) i) A semiconductor is a material with electrical conductivity between that of a conductor (like copper) and an insulator (like glass). Its conductivity can be controlled by doping or temperature.

1. a) ii)

  • A p-type semiconductor is created by doping a pure semiconductor with trivalent impurities (e.g., boron), resulting in a majority of holes as charge carriers.
  • An n-type semiconductor is created by doping a pure semiconductor with pentavalent impurities (e.g., phosphorus), resulting in a majority of electrons as charge carriers.

1. a) iii) An application of a combination of a p-type and n-type semiconductor is a diode (p-n junction).

1. b)

  1. Calculate the total charge (QQ) that flows.

    Q=ItQ = It Q=(8 A)×(4 s)Q = (8 \text{ A}) \times (4 \text{ s}) Q=32 CQ = 32 \text{ C}
  2. Calculate the number of electrons (nn) using the total charge and the charge of a single electron (ee).

    Q=neQ = ne n=Qen = \frac{Q}{e} n=32 C1.6×10−19 Cn = \frac{32 \text{ C}}{1.6 \times 10^{-19} \text{ C}} n=2.0×1020n = 2.0 \times 10^{20}

    The number of electrons is 2.0×1020\boxed{\mathbf{2.0 \times 10^{20}}}.

    2. i) The law of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant if no external forces act on it.

    2. ii)

  3. Identify the mass and velocity of the bullet.

    mbullet=0.05 kgm_{bullet} = 0.05 \text{ kg} vbullet=900 m/sv_{bullet} = 900 \text{ m/s}
  4. Calculate the momentum of the bullet (pbulletp_{bullet}).

    pbullet=mbullet×vbulletp_{bullet} = m_{bullet} \times v_{bullet} pbullet=(0.05 kg)×(900 m/s)p_{bullet} = (0.05 \text{ kg}) \times (900 \text{ m/s}) pbullet=45 kg m/sp_{bullet} = 45 \text{ kg m/s}

    The momentum of the bullet is 45 kg m/s\boxed{\mathbf{45 \text{ kg m/s}}}.

    2. iii)

  5. Apply the law of conservation of momentum.

    The initial momentum of the gun-bullet system is zero.

    mgunugun+mbulletubullet=mgunvgun+mbulletvbulletm_{gun}u_{gun} + m_{bullet}u_{bullet} = m_{gun}v_{gun} + m_{bullet}v_{bullet}

    Since the gun and bullet are initially stationary, ugun=0u_{gun} = 0 and ubullet=0u_{bullet} = 0.

    0=mgunvgun+mbulletvbullet0 = m_{gun}v_{gun} + m_{bullet}v_{bullet}
  6. Rearrange to solve for the recoil velocity of the gun (vgunv_{gun}).

    mgunvgun=−mbulletvbulletm_{gun}v_{gun} = -m_{bullet}v_{bullet} vgun=−mbulletvbulletmgunv_{gun} = -\frac{m_{bullet}v_{bullet}}{m_{gun}} vgun=−(0.05 kg)×(900 m/s)4 kgv_{gun} = -\frac{(0.05 \text{ kg}) \times (900 \text{ m/s})}{4 \text{ kg}} vgun=−45 kg m/s4 kgv_{gun} = -\frac{45 \text{ kg m/s}}{4 \text{ kg}} vgun=−11.25 m/sv_{gun} = -11.25 \text{ m/s}

    The recoil velocity is −11.25 m/s\boxed{\mathbf{-11.25 \text{ m/s}}}. The negative sign indicates the direction opposite to the bullet's velocity.

    3. i)

    • Pressure at point A: Atmospheric pressure
    • Pressure at point B: Pressure of the gas

    3. ii)

  7. Identify the density of mercury (ρ\rho) and the height difference (hh).

    From the diagram, h=0.25 mh = 0.25 \text{ m}.

    ρ=13600 kg/m3\rho = 13600 \text{ kg/m}^3 g=9.81 m/s2(assuming standard gravity)g = 9.81 \text{ m/s}^2 \quad (\text{assuming standard gravity}) h=0.25 mh = 0.25 \text{ m}
  8. Calculate the excess pressure (PexcessP_{excess}) using the formula P=ρghP = \rho gh.

    Pexcess=ρghP_{excess} = \rho gh Pexcess=(13600 kg/m3)×(9.81 m/s2)×(0.25 m)P_{excess} = (13600 \text{ kg/m}^3) \times (9.81 \text{ m/s}^2) \times (0.25 \text{ m}) Pexcess=33354 PaP_{excess} = 33354 \text{ Pa}

    The excess pressure of the gas is 33354 Pa\boxed{\mathbf{33354 \text{ Pa}}}.

    3. iii)

  9. Identify the atmospheric pressure (PatmP_{atm}) and the calculated excess pressure (PexcessP_{excess}).

    Patm=1.0×105 PaP_{atm} = 1.0 \times 10^5 \text{ Pa} Pexcess=33354 PaP_{excess} = 33354 \text{ Pa}
  10. Calculate the pressure of the gas (PgasP_{gas}).

    Pgas=Patm+PexcessP_{gas} = P_{atm} + P_{excess} Pgas=(1.0×105 Pa)+(33354 Pa)P_{gas} = (1.0 \times 10^5 \text{ Pa}) + (33354 \text{ Pa}) Pgas=133354 PaP_{gas} = 133354 \text{ Pa}

    The pressure of the gas is 133354 Pa\boxed{\mathbf{133354 \text{ Pa}}}.

    3. iv) One instrument that is an application of the u-tube is a manometer.

    4. a) For the system to be in equilibrium, two conditions must be met:

    • The net force acting on the system must be zero (translational equilibrium).
    • The net moment (or torque) about any point must be zero (rotational equilibrium).

    4. b)

  11. Apply the principle of moments.

    The sum of clockwise moments about the pivot must equal the sum of anticlockwise moments.
    From the diagram:
    Anticlockwise moment = Force on left ×\times Distance from pivot = 20 N×0.5 m20 \text{ N} \times 0.5 \text{ m}
    Clockwise moment = Force on right ×\times Distance from pivot = 10 N×x10 \text{ N} \times x

  12. Set the moments equal and solve for xx.

    20 N×0.5 m=10 N×x20 \text{ N} \times 0.5 \text{ m} = 10 \text{ N} \times x 10 N m=10 N×x10 \text{ N m} = 10 \text{ N} \times x x=10 N m10 Nx = \frac{10 \text{ N m}}{10 \text{ N}} x=1.0 mx = 1.0 \text{ m}

    The distance xx is 1.0 m\boxed{\mathbf{1.0 \text{ m}}}.

    4. c) i)

  13. Apply the condition for translational equilibrium

    The sum of upward forces equals the sum of downward forces.
    Upward force = RR (at the pivot)
    Downward forces = 20 N+10 N20 \text{ N} + 10 \text{ N}

  14. Solve for RR.

    R=20 N+10 NR = 20 \text{ N} + 10 \text{ N} R=30 NR = 30 \text{ N}

    The upward force, R, acting at the pivot is 30 N\boxed{\mathbf{30 \text{ N}}}.

    4. c) ii)
    A free-body diagram of the forces acting on the see-saw:

    \begin{tikzpicture} % Plank \draw[thick] (-2.5,0) -- (2.5,0); % Pivot \draw[fill=black] (0,-0.2) circle (0.05); \draw (0,-0.2) -- (-0.5,-0.7) -- (0.5,-0.7) -- cycle; % Force R \draw[->, thick] (0, -0.2) -- (0, 0.8) node[above] {$R = 30 \text{ N}$}; % Left load \draw[->, thick] (-1.5, 0) -- (-1.5, -1) node[below] {$20 \text{ N}$}; \draw[fill=gray] (-1.7, 0.1) rectangle (-1.3, 0.3); % Right load \draw[->, thick] (1.0, 0) -- (1.0, -1) node[below] {$10 \text{ N}$}; \draw[fill=gray] (0.8, 0.1) rectangle (1.2, 0.3); % Distances (optional, but good for clarity) \draw[<->] (0, -0.4) -- (-1.5, -0.4) node[midway, below] {$0.5 \text{ m}$}; \draw[<->] (0, -0.4) -- (1.0, -0.4) node[midway, below] {$1.0 \text{ m}$}; \end{tikzpicture}

    The system is stable because it is in equilibrium, meaning the net force and net moment are both zero. This ensures it remains in its current state without accelerating or rotating.

    5. a) Isotopes are atoms of the same element that have the same number of protons (atomic number, Z) but different numbers of neutrons, resulting in different mass numbers (A).

    5. b)

  15. Determine the number of protons (Z) and neutrons (N) for 14^{14}C.

    For Carbon (C), the atomic number is Z=6Z = 6.
    For 14^{14}C, the mass number is A=14A = 14.
    Number of neutrons N=A−Z=14−6=8N = A - Z = 14 - 6 = 8.

  16. Calculate the N/Z ratio.

    NZ=86=43≈1.33\frac{N}{Z} = \frac{8}{6} = \frac{4}{3} \approx 1.33
  17. State if the isotope is stable or unstable.

    For light elements, a stable N/Z ratio is typically close to 1. As the atomic number increases, the stable N/Z ratio increases slightly above 1. For carbon, 12^{12}C has N/Z = 6/6 = 1, which is stable. 14^{14}C has an N/Z ratio of approximately 1.33, which is higher than 1 and indicates an excess of neutrons. Therefore, 14^{14}C is unstable (it is a radioactive isotope that undergoes beta decay).
    The N/Z ratio for 14^{14}C is 1.33\boxed{\mathbf{1.33}}, and the isotope is unstable\boxed{\mathbf{unstable}}.

    5. c) The half-life of a radioactive sample is the time it takes for half of the radioactive nuclei in a sample to decay.

    6. a)

  18. Calculate the equivalent resistance of the two resistors in parallel.

    The 2.0 Ω\Omega and 8.0 Ω\Omega resistors

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