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Charge carriers in the resistor and electrolyte
Apr 10, 2024
Charge carriers in the resistor and electrolyte
Answer
In a resistor, the charge carriers are typically electrons, while in an electrolyte, the charge carriers are ions (cations and anions).
Solution
a
Charge carriers in a resistor: In a metallic conductor such as a resistor, the charge carriers are electrons. These electrons move through the lattice of the metal as a result of an applied electric field, which causes electrical current to flow
b
Charge carriers in an electrolyte: In an electrolyte, which is a liquid or gel that contains ions, the charge carriers are both positive ions (cations) and negative ions (anions). When an electric field is applied, cations move towards the cathode (negative electrode), and anions move towards the anode (positive electrode), resulting in the conduction of electric current
Key Concept
Charge carriers are particles which carry electric charge and are responsible for the conduction of electric current in a material.
Explanation
In resistors, the charge carriers are electrons, while in electrolytes, they are ions. The movement of these charge carriers under the influence of an electric field results in the flow of current.
The half life of fluorine 18 isotope is T. After time t=4T the number of fluorine 18 nuclei in a source is N. How many fluorine 18 nuclei have decayed in the time interval from t=0 to t = 4T
Answer
The number of fluorine-18 nuclei that have decayed from t=0 t = 0 to t=4T t = 4T is N0−N N_0 - N , where N0 N_0 is the initial number of nuclei.
Solution
a
Half-life formula: The number of undecayed nuclei after time t t is given by N=N0(12)tT N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} , where N0 N_0 is the initial number of nuclei, T T is the half-life, and t t is the elapsed time
b
Calculation for t=4T t = 4T : After 4T 4T , the number of undecayed nuclei is N=N0(12)4=N016 N = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}
c
Number of decayed nuclei: The number of decayed nuclei is the initial number minus the remaining number, which is N0−N=N0−N016=15N016 N_0 - N = N_0 - \frac{N_0}{16} = \frac{15N_0}{16} . Since N N is given, we can express N0 N_0 as N0=16N N_0 = 16N
d
Final calculation: Substituting N0 N_0 into the decayed nuclei expression, we get 1516×16N=15N \frac{15}{16} \times 16N = 15N
Key Concept
Radioactive decay and half-life
Explanation
The concept of half-life is used to determine the number of nuclei remaining after a certain period. By knowing the half-life and the elapsed time, we can calculate the fraction of undecayed nuclei and thus find the number of decayed nuclei.
3 3 The activity of an alpha-emitting source is 120kBq120 \mathrm{kBq}. The kinetic energy of each alpha-particle is 4.0 MeV. What is the rate of energy released by the source? A 6.4×10−13 W6.4 \times 10^{-13} \mathrm{~W} B 4.8×10−8 W4.8 \times 10^{-8} \mathrm{~W} C 7.7×10−8 W7.7 \times 10^{-8} \mathrm{~W} D 1.2×105 W1.2 \times 10^{5} \mathrm{~W} Your answer □\square [1] 4 Which of the following statement(s) correctly describe radioactive decay? 1 Radioactive decay can be modelled using dice. 2 Radioactive decay of nuclei is random. 3 Radioactive decay of nuclei is spontaneous. A Only 1 B Only 2 C 2 and 3 D 1,2 and 3 Your answer □\square [1] 5 A gamma-ray photon of frequency 6.76×1022 Hz6.76 \times 10^{22} \mathrm{~Hz} creates a particle-antiparticle pair. The particle-antiparticle pair have zero kinetic energy. What is the mass of the particle? A 2.49×10−28 kg2.49 \times 10^{-28} \mathrm{~kg} B 4.98×10−28 kg4.98 \times 10^{-28} \mathrm{~kg} C 7.47×10−20 kg7.47 \times 10^{-20} \mathrm{~kg} D 4.48×10−11 kg\quad 4.48 \times 10^{-11} \mathrm{~kg}
B
Key Concept
Activity and Energy Release Rate
Explanation
The rate of energy released by a radioactive source is the product of the activity (number of decays per second) and the energy released per decay. The activity is given as 120 kBq, which is 120×103120 \times 10^3 decays per second. Each alpha particle releases 4.0 MeV of energy, which is 4.0×106×1.602×10−194.0 \times 10^6 \times 1.602 \times 10^{-19} joules (since 1 eV = 1.602×10−191.602 \times 10^{-19} joules). The rate of energy release is therefore 120×103×4.0×106×1.602×10−19=7.7×10−8120 \times 10^3 \times 4.0 \times 10^6 \times 1.602 \times 10^{-19} = 7.7 \times 10^{-8} W.
C
Key Concept
Nature of Radioactive Decay
Explanation
Radioactive decay is a random and spontaneous process. It cannot be predicted for individual nuclei but can be modeled statistically, similar to rolling dice. Therefore, statements 2 and 3 are correct descriptions of radioactive decay.
A
Key Concept
Energy-Mass Equivalence
Explanation
The mass of the particle can be found using Einstein's energy-mass equivalence principle, E=mc2E = mc^2. The energy of the gamma-ray photon is E=hνE = h\nu, where hh is Planck's constant and ν\nu is the frequency. Since the photon creates a particle-antiparticle pair with zero kinetic energy, all the photon's energy is converted into mass. Therefore, m=Ec2=hνc2m = \frac{E}{c^2} = \frac{h\nu}{c^2}. Using h=6.626×10−34h = 6.626 \times 10^{-34} J·s and c=3.00×108c = 3.00 \times 10^8 m/s, and the given frequency, the mass of one particle is 2.49×10−282.49 \times 10^{-28} kg.
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4 6 A potential divider circuit is shown below. The battery has electromotive force (e.m.f.) 9.0 V9.0 \mathrm{~V} and negligible internal resistance. At room temperature the potential difference (p.d.) across the thermistor is 4.5 V4.5 \mathrm{~V}. The temperature of the thermistor is increased and its resistance decreases by 20%20 \% from its previous value. What is the p.d. across the thermistor now? A 3.6 V3.6 \mathrm{~V} B 4.0 V4.0 \mathrm{~V} C 5.0 V5.0 \mathrm{~V} D 5.4 V5.4 \mathrm{~V} Your answer [1] 7 A particle is moving at right angles to a uniform magnetic field of flux density BB. The particle has mass mm, charge qq and moves in a circular arc of radius rr in the region of the magnetic field. What quantities are required to determine the momentum of this particle? A B,qB, q and rr B B,qB, q and mm C B,q,rB, q, r and mm D q,rq, r and Your answer [1]
C
Key Concept
Potential Divider Principle
Explanation
When the resistance of the thermistor decreases by 20%, the new resistance is 80% of the original. Since the thermistor and the other resistor form a potential divider, the voltage across the thermistor changes according to the ratio of its resistance to the total resistance. Initially, the thermistor has half the total resistance since it drops half the voltage (4.5V out of 9V). After the decrease, the new voltage across the thermistor can be found using the potential divider equation: Vthermistor=Vtotal×RthermistorRthermistor+RotherV_{\text{thermistor}} = V_{\text{total}} \times \frac{R_{\text{thermistor}}}{R_{\text{thermistor}} + R_{\text{other}}}. Calculating with the new resistance gives a p.d. of 5.0V.
A
Key Concept
Magnetic Force and Circular Motion
Explanation
The momentum of a charged particle moving in a magnetic field can be determined from the magnetic force causing circular motion. The force is given by F=qvBF = qvB, where vv is the velocity of the particle. Since the force is also the centripetal force, F=mv2rF = \frac{mv^2}{r}, equating the two expressions and solving for momentum p=mvp = mv requires knowledge of BB, qq, and rr. Mass mm is not needed as it cancels out when equating the magnetic force to the centripetal force.
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