Question

11 (a) in a tabular form classify the following substances as ferromagnetic or diamagnetic materials: iron, nickel, cobalt, lead, gadolinium, mercury.
(b) two capacitors, 8 μf and 4 μf are connected in parallel and placed in series with a 6 μf capacitor.
(i) draw the circuit diagram for the arrangement.
(ii) calculate the effective capacitance in the circuit.
(c) (i) a step - up transformer is designed to operate from a 20 v supply to deliver 200 v. if the efficiency of the transformer is 80 %. calculate the current in the primary coil when the output terminals are connected to a 300 v and 150 w lamp.
(ii) a charged particle travelling at a speed of 7.0×10^6 ms^(-1) enters at right angle into a magnetic field of strength 0.30 t. if it moves in a circular path of radius 25 cm in the field, calculate the ratio of the charge, q, to the mass, m, of the particle.
12 (a) (i) define decay probability of a radioactive element.
(ii) state the difference between prompt neutrons and delay neutrons.
(b) a metal has a work function of 3.2 ev, calculate:
(i) its threshold frequency;
(ii) the maximum velocity of photoelectrons produced when the substance is illuminated with light of wavelength 4.5×10^(-7) m;
(iii) the stopping potential with light of wavelength 4.5×10^(-7) m.
(c) table 1.0 shows the energy levels of an atom, study it carefully and draw the energy level diagram of the atom.

n	1	2	3	4	5	∞
e_n/ev	-12.30	-5.59	-1.50	-0.75	-0.33	0.00

Explanation:

Step1: Classify substances as magnetic

Use knowledge of magnetic properties of materials.

Step2: Draw capacitor circuit

Follow rules of parallel and series capacitor connections.

Step3: Calculate parallel capacitance

Use \(C_{p}=C_1 + C_2\) for parallel capacitors.

Step4: Calculate series capacitance

Use \(\frac{1}{C_{eff}}=\frac{1}{C_{p}}+\frac{1}{C_3}\) for series capacitors.

Step5: Calculate primary - coil current

Use transformer efficiency and power - voltage relations.

Step6: Calculate charge - to - mass ratio

Equate magnetic and centripetal forces for a charged particle in a magnetic field.

Step7: Define decay probability

State the standard definition.

Step8: Differentiate neutrons

Based on emission time in nuclear fission.

Step9: Calculate threshold frequency

Use \(\phi = h
u_0\).

Step10: Calculate photoelectron velocity

Use photoelectric effect equation \(E=\phi+\frac{1}{2}mv_{max}^{2}\).

Step11: Calculate stopping potential

Relate to photoelectron kinetic energy.

Step12: Draw energy - level diagram

Mark energy levels on a vertical axis.

Answer:

(a)
(i) Ferromagnetic substances: iron, nickel, cobalt; Diamagnetic substances: lead, mercury, gadolinium.
(ii)

1. Circuit - diagram drawing:

• Draw two capacitors \(8\ \mu F\) and \(4\ \mu F\) in parallel (side - by - side with their top plates connected and bottom plates connected). Then connect this parallel combination in series with a \(6\ \mu F\) capacitor.

1. Calculating effective capacitance:

• First, for the parallel combination of \(C_1 = 8\ \mu F\) and \(C_2=4\ \mu F\), the equivalent capacitance of the parallel part \(C_{p}\) is given by \(C_{p}=C_1 + C_2\).
• \(C_{p}=8\ \mu F+4\ \mu F = 12\ \mu F\).
• Now, \(C_{p}=12\ \mu F\) is in series with \(C_3 = 6\ \mu F\). The formula for series capacitance is \(\frac{1}{C_{eff}}=\frac{1}{C_{p}}+\frac{1}{C_3}\).
• Substitute \(C_{p}=12\ \mu F\) and \(C_3 = 6\ \mu F\) into the formula: \(\frac{1}{C_{eff}}=\frac{1}{12}+\frac{1}{6}=\frac{1 + 2}{12}=\frac{3}{12}=\frac{1}{4}\).
• So, \(C_{eff}=4\ \mu F\).

(b)
(i) Let \(V_p = 20\ V\), \(V_s=200\ V\), and \(\eta = 80\%=0.8\). Let the power in the secondary coil be \(P_s\). Assume the power in the primary coil is \(P_p\). Since \(\eta=\frac{P_s}{P_p}\), and for an ideal - like case (using the transformer voltage - current relation \(V_pI_p = V_sI_s\)). Let the power in the secondary coil be used to power a \(150\ W\) lamp. So \(P_s = 150\ W\). Then \(P_p=\frac{P_s}{\eta}=\frac{150}{0.8}=187.5\ W\). Using \(P_p = V_pI_p\), we can find \(I_p=\frac{P_p}{V_p}=\frac{187.5}{20}=9.375\ A\).
(ii) Given \(v = 7.0\times10^{6}\ m/s\), \(B = 0.30\ T\), and \(r = 25\ cm=0.25\ m\). When a charged particle moves in a circular path in a magnetic field, the magnetic force \(F = qvB\) provides the centripetal force \(F_c=\frac{mv^{2}}{r}\). So \(qvB=\frac{mv^{2}}{r}\), and \(\frac{q}{m}=\frac{v}{Br}\).

• Substitute the values: \(\frac{q}{m}=\frac{7.0\times10^{6}}{0.30\times0.25}=\frac{7.0\times10^{6}}{0.075}\approx9.33\times10^{7}\ C/kg\).

(c)
(i) The decay probability of a radioactive element is the probability that a given radioactive nucleus will decay in a given time interval.
(ii) Prompt neutrons are emitted almost instantaneously (within \(10^{-14}\ s\)) during a nuclear fission event, while delayed neutrons are emitted some time (from milliseconds to minutes) after the fission event.
(d)
(i) Given the work - function \(\phi=3.2\ eV = 3.2\times1.6\times10^{-19}\ J\). Using the formula \(\phi = h
u_0\), where \(h = 6.63\times10^{-34}\ Js\), \(
u_0=\frac{\phi}{h}=\frac{3.2\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx7.7\times10^{14}\ Hz\).
(ii) Given \(\lambda = 4.5\times10^{-7}\ m\), the energy of the incident photon is \(E = h\frac{c}{\lambda}\), where \(c = 3\times10^{8}\ m/s\). \(E=\frac{6.63\times10^{-34}\times3\times10^{8}}{4.5\times10^{-7}}=4.42\times10^{-19}\ J\). Using the photoelectric effect equation \(E=\phi+\frac{1}{2}mv_{max}^{2}\), \(\frac{1}{2}mv_{max}^{2}=E - \phi\). First, convert \(\phi = 3.2\ eV=3.2\times1.6\times10^{-19}\ J = 5.12\times10^{-19}\ J\). \(E - \phi=4.42\times10^{-19}-5.12\times10^{-19}=- 0.7\times10^{-19}\ J\) (This is wrong as \(E<\phi\), no photoelectrons are emitted, \(v_{max}=0\)).
(iii) Since \(E<\phi\), no photoelectrons are emitted, so the stopping potential \(V_0 = 0\ V\).
(e)
Draw a vertical axis labeled 'Energy (eV)' and a horizontal axis with no specific label. Mark the energy levels \(E_0 = 0\ eV\), \(E_1=- 12.30\ eV\), \(E_2=-5.59\ eV\), \(E_3=-1.50\ eV\), \(E_4=-0.75\ eV\), \(E_5=-0.33\ eV\) as horizontal lines on the vertical - energy axis. Label each line with the corresponding \(n\) value (\(n = 0,1,2,3,4,5\)) and the energy value.