11 (a) in a tabular form classify the following substances as ferromagnetic or diamagnetic materials: iron…

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

Answer

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)