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 m s^(-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: Analyze capacitor - connection problem

For two capacitors \(C_1 = 8\ \mu F\) and \(C_2=1\ \mu F\) in parallel, the equivalent capacitance \(C_{p}\) of parallel - connected capacitors is given by \(C_{p}=C_1 + C_2\).
\[C_{p}=8\ \mu F+ 1\ \mu F=9\ \mu F\]
This parallel combination is in series with \(C_3 = 6\ \mu F\). The formula for the equivalent capacitance \(C_{eq}\) of two capacitors \(C_{p}\) and \(C_3\) in series is \(\frac{1}{C_{eq}}=\frac{1}{C_{p}}+\frac{1}{C_{3}}\).
\[ \frac{1}{C_{eq}}=\frac{1}{9\ \mu F}+\frac{1}{6\ \mu F}=\frac{2 + 3}{18\ \mu F}=\frac{5}{18\ \mu F}\]
\[C_{eq}=\frac{18}{5}\ \mu F = 3.6\ \mu F\]

Step2: Analyze transformer problem

The power input \(P_{in}\) and power output \(P_{out}\) are related by the efficiency formula \(\eta=\frac{P_{out}}{P_{in}}\). Given \(\eta = 0.8\), \(V_{out}=200\ V\). Let the current in the secondary coil be \(I_{s}\), and assume the power output is used to power a \(P = 150\ W\) lamp. So \(P_{out}=150\ W\). Then \(P_{in}=\frac{P_{out}}{\eta}=\frac{150\ W}{0.8}=187.5\ W\). Also, \(P_{in}=V_{in}I_{in}\), where \(V_{in}=20\ V\). So \(I_{in}=\frac{P_{in}}{V_{in}}=\frac{187.5\ W}{20\ V}=9.375\ A\)

Step3: Analyze charged - particle in magnetic - field problem

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 the ratio \(\frac{q}{m}=\frac{v}{Br}\).
Given \(v = 7.0\times10^{6}\ m/s\), \(B = 0.30\ T\), \(r = 25\ cm=0.25\ m\)
\[\frac{q}{m}=\frac{7.0\times10^{6}\ m/s}{0.30\ T\times0.25\ m}=\frac{7.0\times10^{6}}{0.075}\ C/kg\approx9.33\times10^{7}\ C/kg\]

Step4: Analyze photoelectric - effect problem

(i) The work - function \(\phi\) and the threshold frequency \(f_0\) are related by \(\phi = hf_0\), where \(h = 6.63\times10^{-34}\ J\cdot s\) and \(\phi=3.2\ eV=3.2\times1.6\times10^{-19}\ J = 5.12\times10^{-19}\ J\)
\[f_0=\frac{\phi}{h}=\frac{5.12\times10^{-19}\ J}{6.63\times10^{-34}\ J\cdot s}\approx7.72\times10^{14}\ Hz\]
(ii) First, find the energy of the incident photon \(E = h\frac{c}{\lambda}\), where \(c = 3\times10^{8}\ m/s\), \(\lambda=4.5\times10^{-7}\ m\).
\[E=h\frac{c}{\lambda}=\frac{6.63\times10^{-34}\ J\cdot s\times3\times10^{8}\ m/s}{4.5\times10^{-7}\ m}=4.42\times10^{-19}\ J\]
The maximum kinetic energy of the photoelectrons \(K_{max}=E-\phi\).
\[K_{max}=4.42\times10^{-19}\ J - 5.12\times10^{-19}\ J=- 0.7\times10^{-19}\ J\] (This is wrong, because \(E<\phi\) for this part, there will be no photoelectric effect. Let's assume the work - function is wrong or the wavelength is wrong. If we assume the correct work - function and wavelength values, \(K_{max}=E - \phi\), and \(K_{max}=\frac{1}{2}mv_{max}^{2}\), so \(v_{max}=\sqrt{\frac{2K_{max}}{m}}\))
(iii) The stopping potential \(V_0\) is related to the maximum kinetic energy of the photoelectrons by \(K_{max}=eV_0\)

Answer:

• For the capacitor problem, the effective capacitance is \(3.6\ \mu F\)
• For the transformer problem, the current in the primary coil is \(9.375\ A\)
• For the charged - particle in magnetic - field problem, \(\frac{q}{m}\approx9.33\times10^{7}\ C/kg\)
• For the photoelectric - effect problem (i), the threshold frequency is approximately \(7.72\times10^{14}\ Hz\)