JEE > I. E. Irodov Solutions for Physics Class 11 & Class 12 > Irodov Solutions: Radioactivity- 2

1 Crore+ students have signed up on EduRev. Have you? |

**Q.231. A radionuclide A _{1} goes through the transformation chain A_{1 }→ A_{2} → A_{3 } (stable) with respective decay constants λ_{1} and λ_{2}. Assuming that at the initial moment the preparation contained only the radionuclide A_{1} equal in quantity to N_{10} nuclei, find the equation describing accumulation of the stable isotope A_{3}.**

**Ans.** Here we have the equations

From problem 229

Then

or

since N_{3} = 0 initially

So

**Q.232. A Bi21° radionuclide decays via the chain **

where the decay constants are 5.80.10^{-8} s^{-1}. Calculate alpha- and beta-activities of the Bi^{210} preparation of mass 1.00 mg a month after its manufacture.

**Ans** We have the chain

of the previous problem initially

A month after preparation

N_{1 =} 4.54 x 10^{16 }

N_{2} = 2.52 x 10^{18}

using the results of the previous problem.

Then

Aβ = λ_{1}N_{1} = 0.725 x 10^{11 }dis/sec

Aα = λ_{2}N_{2} = 1.46 x 10^{11} dis/sec

**Q.233. (a) What isotope is produced from the alpha-radioactive Ra ^{228} as a result of five alpha-disintegrations and four β-disintegrations? (b) How many alpha- and β-decays does U^{ 238 } experience before turning finally into the stable Pb^{206 } isotope? **

**Ans. ** (a) Ra has Z - 88, A - 226 After 3 a emission and 4 p (electron) emission

A - 206

Z - 88 + 4-5x2 - 82

The product is ^{82}pb^{206 }

(b) We require

- ΔZ = 10 = 2 n - m

- ΔA = 32 = ft x 4

Here n = no. of α emissions

m = no. of β emissions

Thus n = 8, m = 6

**Q.234. A stationary Pb ^{200} nucleus emits an alpha-particle with kinetic energy T_{α} = 5.77 MeV. Find the recoil velocity of a daughter nucleus. What fraction of the total energy liberated in this decay is accounted for by the recoil energy of the daughter nucleus? **

**Ans. **The momentum of the α-particle is

This is also the recoil m om entum o f the daughter nuclear in opposite direction. The recoil velocity of the daughter ntideus is

= 3.39x10^{5} m/S

The eneigy of the daughter nucleus is and this represents a fraction

of total eneigy. Here Md is the mass of the daughter nudeus.

**Q.235. Find the amount of heat generated by 1.00 mg of a Po ^{210} preparation during the mean lifetime period of these nuclei if the emitted alpha-particles are known to possess the kinetic energy 5.3 MeV and practically all daughter nuclei are formed directly in the ground state.**

**Ans. **The number of nuclei initially present is

In the mean life time of these nuclei the number decaying is the fraction Thus the eneigy released is 2.87 x 10^{18} x 0.632 x 5.3 x 1.602 x 10^{-13} J = 1.54MJ

**Q.236. The alpha-decay of Po ^{210} nuclei (in the ground state) is accompanied by emission of two groups of alpha-particles with kinetic energies 5.30 and 4.50 MeV. Following the emission of these particles the daughter nuclei are found in the ground and excited states. Find the energy of gamma-quanta emitted by the excited nuclei.**

**Ans.** We neglect all recoil effects. Then the following diagram gives the eneigy of the gamma ray

**Q.237. The mean path length of alpha-particles in air under standard conditions is defined by the formula R = 0.98.10 ^{-27} v^{3}_{o} cm, where v_{0} (cm/s) is the initial velocity of an alpha-particle. Using this formula, find for an alpha-particle with initial kinetic energy 7.0 MeV:**

(a) its mean path length;

(b) the average number of ion pairs formed by the given alphaparticle over the whole path R as well as over its first half, assuming the ion pair formation energy to be equal to 34 eV.

**Ans. ** (a) For an alpha particle with initial K.E. 7.0 MeV, the initial velocity is

= 1.83 x 10^{9}an/sec

Thus R = 6.02an

(b) Over the whole path the number of ion pairs is

Over the first half of the path We write the formula for the mean path as RαE^{1/2 }where E is the initial energy. Thus if the energy of the a-particle after traversing the first half of the path is E_{1 }then

Hence number of ion pairs formed in the first half of the path length is

**Q.238. Find the energy Q liberated in β ^{-}- and β^{+}-decays and in K-capture if the masses of the parent atom M_{P}, the daughter atom M_{d} and an electron m are known.**

**Ans. ** In β^{-} decay

since M_{p}, M_{d} are the masses of the atoms. The binding energy of the electrons in ignored. In K capture

In β^{+} decay

Then

**Q.239. Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta-particles emitted by Be ^{10} nuclei and the corresponding kinetic energy of recoiling daughter nuclei formed directly in the ground state. **

**Ans.** The reaction is

For maximum K.E. of electrons we can put the energy of to be zero. The atomic masses are

Be^{10 =} 10.016711 amu

B^{10} = 10.016114 amu

So the K.E. of electrons is (see previous problem)

597 x 10^{-6} amu x c^{2} = 0.56 MeV

The momentum of electrons with this K.E. is 0.941

and the recoil energy of the daughter is

**Q.240. Evaluate the amount of heat produced during a day by a **β^{-}**-active Na ^{24} preparation of mass m = 1.0 mg. The beta-particles are assumed to possess an average kinetic energy equal to 1/3 of the highest possible energy of the given decay. The half-life of Na^{24} is T = 15 hours. **

**Ans. **The masses are

Na^{24 }= 24 - 0.00903 amu and Mg^{24} = 24 - 0.01496 amu

The reaction is

The maximum K.E. of electrons is

0.00593 x 93 MeV = 5.52 MeV

Average K.E. according to the problem is then = 1.84 MeV

The initial number of Na^{24} is

The fraction decaying in a day is

1 - (2) ^{- 24/15} = 0.67

Hence the heat produced in a day is

0.67 x 2.51 x 10^{19 }x 1.84 x 1.602 x 10^{-13} Joul = 4.95MJ

**Q.241. Taking the values of atomic masses from the tables, calculate the kinetic energies of a positron and a neutrino emitted by C ^{11} nucleus for the case when the daughter nucleus does not recoil. **

**Ans. ** We assume that the parent nucleus is at rest Then since the daughter nucleus does not recoil, we have

i.e. positron & v mometum are equal and opposite. On the other hand

total energy released. (Here we have used the fact that energy

of the neutrino is

Now

Then

Thus c p = 0.646 M eV = energy of neutrino

Also K.E. of electron = 1.47 - 0.646 - 0.511 = 0.313 MeV

**Q.242. Find the kinetic energy of the recoil nucleus in the positronic decay of a Nn nucleus for the case when the energy of positrons is maximum. **

**Ans. **The K.E. of the positron is maximum when the energy of neutrino is zero. Since the recoil energy of the nucleus is quite small, it can be calculated by successive approximation.

The reaction is

The maximum energy available to the positron (including its rest energy) is

c^{2} (Mass of N^{13 }nucleus - Mass of C^{13} nucleus)

= c^{2} (Mass of N^{13} atom - Mass of C^{13} atom - m_{e})

= 0.00239 c^{2} - m_{e}c^{2}

= (0.00239 x 931 - 0.511) MeV

= 1.71 MeV

The momentum corresponding to this energy is 1.636 MeV/c .

The recoil energy of the nucleus is then

= 111 eV = 0.111 keV

on using Mc^{2} - 13 x 931 MeV

**Q.243. From the tables of atomic masses determine the velocity of a nucleus appearing as a result of K-capture in a Bel atom provided the daughter nucleus turns out to be in the ground state. **

**Ans. **The process is

The energy available in the process is

Q = c^{2} (Mass of Be^{7} atom - Mass of Li^{-7} atom)

= 0.00092 x 931 MeV = 0.86 MeV

The momentum of a K electron is negligible. So in the rest frame of the Be^{7} atom, most of the energy is taken by neutrino whose momentum is very nearly 0.86MeV/c The momentum of the recoiling nucleus is equal and opposite. The velocity of recoil is

= 3.96 x10^{6} cm/s

**Q.244. Passing down to the ground state, excited Ag ^{109} in nuclei emit either gamma quanta with energy 87 keV or K conversion electrons whose binding energy is 26 keV. Find the velocity of these electrons. **

**Ans. ** In internal conversion, the total energy is used to knock out K electrons. The KE. of these electrons is energy available-B.E. of K electrons

= (87 - 26) = 61 keV

The total energy including rest mass of electrons is 0.511 + 0.061 = 0.572 MeV

The momentum corresponding to this total energy is

0257 MeV/c.

The velocity is then =

**Q.245. A free stationary Ir ^{191} nucleus with excitation energy E = 129 keV passes to the ground state, emitting a gamma quantum. Calculate the fractional change of gamma quanta energy due to recoil of the nucleus. **

**Ans. **With recoil neglected, the y-quantram will have 129 keV eneigy. To a first approximation, its momentrum will be 129 keV/c and the energy of recoil will be

= 4.18 x 10^{- 8} MeV

In the next approximation we therefore write

Therefore

**Q.246. What must be the relative velocity of a source and an absorber consisting of free Ir ^{191} nuclei to observe the maximum absorption of gamma quanta with energy ε = 129 keV? **

**Ans. **For maximum (resonant) absorption, the absorbing nucleus must be moving with enough speed to cancel the momentum of the oncoming photon and have just right eneigy (ε = 129 keV) available for transition to the excited state.

Since and momentum of the photon is these condition can be satisfied if the velocity of the nucleus is

218m/s = 0.218km/s

**Q.247. A source of gamma quanta is placed at a height h = 20m above an absorber. With what velocity should the source be displaced upward to counterbalance completely the gravitational variation of gamma quanta energy due to the Earth's gravity at the point where the absorber is located? **

**Ans. **Because of the gravitational shift the frequency of the gamma ray at the location of the absorber is increased by

For this to be compensated by the Doppler shift (assuming that resonant absorption is possible in the absence of gravitational field) we must have

**Q.248. What is the minimum height to which a gamma quanta source containing excited Zn ^{67 } nuclei has to be raised for the gravitational displacement of the Mossbauer line to exceed the line width itself, when registered on the Earth's surface? The registered gamma quanta are known to have an energy ε = 93 keV and appear on transition of Zn^{67} nuclei to the ground state, and the mean lifetime of the excited state is ζ = 14μS.**

**Ans. **The natural life time is

Thus the condition implies

= 4.64 metre

(h here is height of the place, not planck’s constant.)

The document Irodov Solutions: Radioactivity- 2 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

All you need of JEE at this link: JEE

Download as PDF