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3.20 b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (iii) 6 Ω

1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.

Q. 13.31     Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which  was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.

Let the amount of Energy to be produced using nuclear power per year in 2020 is E      (Only 10% of the required electrical energy is to be produced by Nuclear power and only 25% of                                                                                                         thermo-nuclear is successfully converted into electrical energy) Amount of Uranium required to produce this...

Q. 13.30     Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.

(a)   The above fusion reaction releases the energy of 26 MeV Number of Hydrogen atoms in 1.0 kg of Hydrogen is 1000NA Therefore 250NA such reactions would take place The energy released in the whole process is E1 (b) The energy released in fission of one  atom is 200 MeV Number of  atoms present in 1 kg of  is N The energy released on fission of N atoms is E2

Q. 13.29     Obtain the maximum kinetic energy of $\beta$- particles, and the radiation frequencies of $\gamma$ decays in the decay                    scheme shown in Fig. 13.6. You are given that

$m(^{198}Au)=197.968233\; u$

$m(^{198}Hg)=197.966760 \; u$

decays from 1.088 MeV to 0 V Frequency of  is  Plank's constant, h=6.6210-34 Js  Similarly, we can calculate frequencies of  and  The energy of the highest level would be equal to the energy released after the decay Mass defect is We know 1u = 931.5 MeV/c2 Q value= 0.001473931.5=1.3721 MeV The maximum Kinetic energy of  would be 1.3721-1.088=0.2841 MeV The maximum Kinetic energy of  would be...

Q.13.28     Consider the D–T reaction (deuterium–tritium fusion)

$_{1}^{2}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+n$

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles $= 2(3kT/2)$; k = Boltzman’s constant, T = absolute temperature.)

To initiate the reaction both the nuclei would have to come in contact with each other. Just before the reaction the distance between their centres would be 4.0 fm. The electrostatic potential energy of the system at that point would be The same amount of Kinetic Energy K would be required to overcome the electrostatic forces of repulsion to initiate the reaction It is given that  Therefore...

Q.13.28     Consider the D–T reaction (deuterium-tritium fusion)

$_{1}^{2}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+n$

(a) Calculate the energy released in MeV in this reaction from the data:

$m(_{1}^{2}\textrm{H})=2.014102\; u$

$m(_{1}^{3}\textrm{H})=3.016049\; u$

The mass defect of the reaction is  1u = 931.5 MeV/c2 Q=0.018883931.5=17.59 MeV

Q.13.27     Consider the fission of $_{92}^{238}\textrm{U}$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are $_{58}^{140}\textrm{Ce}$ and $_{44}^{99}\textrm{Ru}$. Calculate Q   for this fission process. The relevant atomic and particle masses are

$m(_{92}^{238}\textrm{U})=238.05079\; u$

$m(_{58}^{140}\textrm{Ce})=139.90543\; u$

$m(_{44}^{99}\textrm{Ru})= 98.90594\; u$

The fission reaction given in the question can be written as The mass defect for the above reaction would be In the above equation, mN represents nuclear masses but 1u =931.5 MeV/c2 Q=0.247995931.5 Q=231.007 MeV Q value of the fission process is 231.007 MeV

Q.13.26     Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha$-particle. Consider the following decay processes:

$_{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}$

$_{88}^{223}\textrm{Ra}\rightarrow _{86}^{219}\textrm{Rn}+_{2}^{4}\textrm{He}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

1 u = 931.5 MeV/c2 Q=0.03419931.5 =31.848 MeV As the Q value is positive the reaction is energetically allowed 1 u = 931.5 MeV/c2 Q=0.00642931.5 =5.98 MeV As the Q value is positive the reaction is energetically allowed

Q.13.25     A source contains two phosphorous radio nuclides $_{15}^{32}\textrm{P}(T_{1/2}=14.3d)$ and $_{15}^{33}\textrm{P}(T_{1/2}=25.3d)$. Initially, 10% of the decays come from $_{15}^{33}\textrm{P}$. How long one must wait until 90% do so?

Let initially there be N1 atoms of  and N2 atoms of  and let their  decay constants be  and  respectively Since initially the activity of   is 1/9 times that of  we have       (i) Let after time t the activity of   be 9 times that of      (ii) Dividing equation (ii) by (i) and taking the natural log of both sides we get where  and t comes out to be 208.5 days

Q.13.24     The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $_{13}^{27}\textrm{Al}$ from the following data:

$m(_{20}^{40}\textrm{Ca})=39.962591\; u$

$m(_{20}^{41}\textrm{Ca})=40.962278 \; u$

$m(_{13}^{26}\textrm{Al})=25.986895 \; u$

$m(_{13}^{27}\textrm{Al})=26.981541 \; u$

The reaction showing the neutron separation is But 1u=931.5 MeV/c2 Therefore E=(0.014019)931.5 E=13.059 MeV Therefore to remove a neutron from the  nucleus 13.059 MeV of energy is required

Q.13.24     The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $_{20}^{41}\textrm{Ca}$ from the following data:

$m(_{20}^{40}\textrm{Ca})=39.962591\; u$

$m(_{20}^{41}\textrm{Ca})=40.962278 \; u$

$m(_{13}^{26}\textrm{Al})=25.986895 \; u$

$m(_{13}^{27}\textrm{Al})=26.981541 \; u$

The reaction showing the neutron separation is But 1u=931.5 MeV/c2 Therefore E=(0.008978)931.5 E=8.363007 MeV Therefore to remove a neutron from the  nucleus 8.363007 MeV of energy is required

Q.13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are  $_{12}^{24}\textrm{Mg}(23.98504\; u)$$_{12}^{25}\textrm{Mg}(24.98584\; u)$ and $_{12}^{26}\textrm{Mg}(25.98259\; u)$. The natural abundance of is 78.99% by mass. Calculate the abundances of other two isotopes.

Let the abundances of  and  be x and y respectively. x+y+78.99=100 y=21.01-x The average atomic mass of Mg is 24.312 u The abundances of  and   are 9.3% and 11.71% respectively

Q : 11.37  Answer the following questions:

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
$\dpi{100} E=hv,p=\frac{h}{\lambda }$
But while the value of $\dpi{100} \lambda$ is physically significant, the value of  $\dpi{100} v$  (and therefore, the value of the phase speed $\dpi{100} v \lambda$ ) has no physical significance. Why?

The absolute energy has no significance because of the reference point being arbitrary and thus the inclusion of an arbitrary constant rendering the value of and . to have no physical significance as such. The group speed is defined as   Due to the significance of the group speed the absolute value of wavelength has physical significance.

Q : 11.37        Answer the following questions:

(d)  Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

The work function is defined as the minimum energy below which an electron will never be ejected from the metal. But when photons with high energy are incident it is possible that electrons from different orbits get ejected and would therefore come out of the atom with different kinetic energies.

Q : 11.37        Answer the following questions:

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

At ordinary pressure due to a large number of collisions among themselves, the gases have no chance of reaching the electrodes while at very low pressure these collisions decrease exponentially and the gas molecules have a chance of reaching the respective electrodes and therefore are capable of conducting electricity.

Q : 11.37       Answer the following questions:

(b) What is so special about the combination $\dpi{100} e/m$ ? Why do we not simply talk of $\dpi{100} e$ and $\dpi{100} m$ separately?

The speed of a charged particle is given by the relations or As we can see the speed depends on the ratio e/m it is of such huge importance.

Q : 11.37        Answer the following questions:

(a) Quarks inside protons and neutrons are thought to carry fractional charges     $\dpi{100} [(+2/3)e;(-1/3)e]$. Why do they not show up in Millikan’s oil-drop experiment?

Quarks are thought to be tight within a proton or neutron by forces which grow tough if one tries to pull them apart. That is event though fractional charges may exist in nature, the observable charges are still integral multiples of the charge of the electron (e)

Q : 11.36  Compute the typical de Broglie wavelength of an electron in a metal at $\dpi{100} 27\hspace{1mm}^{\circ}C$ and compare it with the mean separation between two electrons in a metal which is given to be about  $\dpi{100} 2\times 10^-^1^0\hspace{1mm}m$.

The de Broglie wavelength associated with the electrons is The de Broglie wavelength of the electrons is comparable to the mean separation between two electrons.

Q : 11.35     Find the typical de Broglie wavelength associated with a $He$ atom in helium gas at room temperature ($\dpi{100} 27\hspace{1mm}^{\circ}C$) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

The kinetic energy K of a He atom is given by mHe i.e. mass of one atom of He can be calculated as follows                              (NA is the Avogadro's Number) De Broglie wavelength is given by Mean separation between two atoms is given by the relation From the ideal gas equation we have The mean separation is therefore The mean separation is greater than the de Broglie wavelength.

Q : 11.34    The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale  of  $\dpi{100} 10^-^1^5\hspace{1mm}m$ or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams.

(Rest mass energy of electron $\dpi{100} =0.511\hspace{1mm}MeV$.)

Rest mass of the electron  Momentum  using relativistic formula for energy
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