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.\n(b) two capacitors, 8 μf and 4 μf are connected in parallel and placed in series with a 6 μf capacitor.\n(i) draw the circuit diagram for the arrangement.\n(ii) calculate the effective capacitance in the circuit.\n(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.\n(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.\n12 (a) (i) define decay probability of a radioactive element.\n(ii) state the difference between prompt neutrons and delay neutrons.\n(b) a metal has a work function of 3.2 ev, calculate:\n(i) its threshold frequency;\n(ii) the maximum velocity of photoelectrons produced when the substance is illuminated with light of wavelength 4.5×10^(-7) m;\n(iii) the stopping potential with light of wavelength 4.5×10^(-7) m.\n(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|

11 (a) in a tabular form classify the following substances as ferromagnetic or diamagnetic materials: iron, nickel, cobalt, lead, gadolinium, mercury.\n(b) two capacitors, 8 μf and 4 μf are connected in parallel and placed in series with a 6 μf capacitor.\n(i) draw the circuit diagram for the arrangement.\n(ii) calculate the effective capacitance in the circuit.\n(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.\n(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.\n12 (a) (i) define decay probability of a radioactive element.\n(ii) state the difference between prompt neutrons and delay neutrons.\n(b) a metal has a work function of 3.2 ev, calculate:\n(i) its threshold frequency;\n(ii) the maximum velocity of photoelectrons produced when the substance is illuminated with light of wavelength 4.5×10^(-7) m;\n(iii) the stopping potential with light of wavelength 4.5×10^(-7) m.\n(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

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.
  2. 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\nu_0), where (h = 6.63\times10^{-34}\ Js), (\nu_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.

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\nu_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.