Q: The number of disintegrations per second that will occur in 1 gram of ^{238}_{92}U is:(if its half-life against \alpha decay is 1.42\times 10^{17}s )

t_{1/2} = 1.42\times 10^{17} s 1\ gm of U contains =\dfrac{6\times 10^{23}}{235} =25.3\times 10^{20} atomsThe decay rate: R = \lambda N = \dfrac{ln \ 2}{t_{1/2}}\times 25.3\times 10^{20} = \dfrac{.693}{1.42\times 10^{17}}\times 25.3\times 10^{20} =0.1235\times{ 10 }^{ 5 }{ s }^{ -1 }

Q: If the decay constant of a radioactive nuclide is 6.93 x 10 ^{-3} sec ^{-1} , its half life in minutes is:

\lambda = 6.93\times 10^{-3} {sec}^{-1} {t}_{1/2} = \dfrac{.693}{.693\times 10^{-2}} sec = \dfrac{100}{60} min = 1.67 min

Question 1

The nuclide   ^{14} C  contains (a) how many protons and (b) how many neutrons?

Accepted Answer

(a) 6 protons, since  Z = 6  for carbon (b)8 neutrons, since  A &#8211; Z = 14 &#8211; 6 = 8  .

Question 2

During   \beta^{-}   decay, a neutron inside nucleus converts into proton, electron and  x . Then, what is the particle  x ?

Accepted Answer

Process of  \beta  decay         n^{0} \rightarrow  _{1}P + e^{-} + \bar{V}e where,  p = proton                e^{-} = electron               \bar{V}e = anti \  neutrino

Question 3

Question 4

Explain briefly the process of archaeological dating?

Accepted Answer

Carbon, is a naturally occurring isotope of the element carbon. When a unrg thing dies, it stops interacting with biosphere, and the carbon 14 in it remains unaffected by the biosphere but will it naturally undergoes decay. Decay of carbon 14 takes thousands of years and it is their wonder of nature that form basis of radioactive dating and made this carbon 14 analysis of powerful tool. The process of radiocarbon dating starts with the analysis of carbon 14 left in the sample. The proportion of carbon 14 left in the sample examined provides an indication of the time elapsed since the death of the sample source

Question 5

Give any two important sources of background radiations.

Accepted Answer

Two sources of background radiations :  K-40   and  C-14

Question 6

Find the energy liberated in the reaction

^{223} R a \to^{209} P b +^{14} C

the atomic masses needed are as follows

^{223} R a^{209} P b^{14} C

22.018 u 208.981 u 14.003 u

View solution

Accepted Answer

^{223}R_a = 223.018 u, ^{209}Pb= 208.981 u; ^{14}C = 14.003 u   ^{223}R_a ightarrow ^{209}Pb + ^{14}C   	riangle m = mass ^{223}R_a - mass ( ^{209}Pb +^{14}C )   ightarrow = 223.018 -(208.981 +14.003) = 0.034   energy =  	riangle M 	imes u = 0.034 	imes 931 = 31.65 MeV

Question 7

The number of disintegrations per second that will occur in  1  gram of  ^{238}_{92}U  is:(if its half-life against  \alpha   decay is  1.42	imes 10^{17}s )

Accepted Answer

t_{1/2} = 1.42	imes 10^{17} s 1\ gm  of U contains  =\dfrac{6	imes 10^{23}}{235}                                  =25.3	imes 10^{20}  atomsThe decay rate: R = \lambda N    = \dfrac{ln \ 2}{t_{1/2}}	imes 25.3	imes 10^{20}      = \dfrac{.693}{1.42	imes 10^{17}}	imes 25.3	imes 10^{20}      =0.1235	imes{ 10 }^{ 5 }{ s }^{ -1 }

Question 8

In the nuclear decay given below; \displaystyle ^A_ZX \rightarrow ^A_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B* \rightarrow ^{A-4}_{Z-1}B The particles emitted in the sequence are:

Accepted Answer

\cdot \alpha - decay           { Z }^{ { X }^{ A } }\xrightarrow { \alpha  } { Z-2 }^{ { X }^{ A-4 } } \cdot \beta - decay           { Z }^{ { X }^{ A } }\xrightarrow { \beta  } { Z+1 }^{ { X }^{ A } } \cdot \gamma - decay consists of electromagnetic radioations.Now, the given decay series.           { Z }^{ { X }^{ A } }\quad \underrightarrow { \beta  } \quad { Z+1 }^{ { Y }^{ A } }\quad \underrightarrow { \alpha  } \quad { Z-1 }^{ { B }^{ A-4 } }\quad \underrightarrow { \gamma  } \quad { Z-1 }^{ { B }^{ A-4 } } 	herefore   The correct sequence.           \beta , \alpha , \gamma

Question 9

If the decay constant of a radioactive nuclide is  6.93  x  10 ^{-3} sec ^{-1} , its half life in minutes is:

Accepted Answer

\lambda = 6.93	imes 10^{-3}   {sec}^{-1} {t}_{1/2} = \dfrac{.693}{.693	imes 10^{-2}} sec   = \dfrac{100}{60} min   = 1.67 min

Question 10

If  x  gm of a nuclear fuel of mass number  A  undergoes fission inside a reactor, then the number of fission will be ( N -Avogadro number)

Accepted Answer

Power of a nuclear reactor :The output power of a nuclear reactor is given by: P = nE       ----(1)Where,  n =  No. of fission taking place in 1 sec             E =   Energy released per fission.If  'x'  grams of a nuclear fuel of mass number 'A' undergo fission in a time of  't'  sec and 'E'  is energy released per fission, then the power output of the nuclear reactor is given by: P = \dfrac{NEx}{At}    --(2) where N - Avagadro number.Comparing equation (1) and (2) we get: n = \dfrac{Nx}{At} Where  n  is no. of fission taking place in  1  sec.So, for ' t ' sec time, no. of fission on ' t ' sec will be \dfrac{Nx}{At}t = \dfrac{Nx}{A} So, correct choice is option C.

Question 11

Differentiate between half life and average life of a radioactive substance.

Accepted Answer

The difference between half life and average life is given below :- Half lifeAverage life (i) The time required for half of the radioactive substance to decay is called as half life of that substance.(ii) It is denoted by  t _{\frac{1}{2}} (iii)  t_{\frac{1}{2}} =   half life  = \left(\dfrac{\ln 2}{\lambda} ight) where  \lambda  is decay constant.(i) The expected life for the radioactive substance within a sample is called as average life of that substance.(ii) It is denoted by  	au .(iii)  	au =  average life  = \dfrac{1}{\lambda}

Question 12

From the relation R =  R_0A^\frac{1}{3}  , where  R_0  is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Accepted Answer

Density of nuclear matter  =  Mass of nucleus/Volume of nucleus                                             =\dfrac{A}{\dfrac{4}{3}\pi R^3}                                             =\dfrac{A}{\dfrac{4}{3}\pi (R_{0}A^{1/3})^3}\ =\dfrac{1}{\dfrac{4}{3}\pi R_0^3}=constant

Question 13

The mass of deutron  (_1H^2)  nucleus is  2.013553  a.m.u. If the masses of proton and neutron are  1.007275  a.m.u. and  1.008665  a.m.u. respectively, calculate the mass defect, the packing fraction , binding energy and binding energu per nucleon.

Accepted Answer

Mass defect  =  Mass of atom  -  Mass of neutron and protonMass defect  =2.013553-(1.007275+1.008665) \=0.002387 Binding Energy  =0.002387	imes 931=2.22MeV Binding Energy per nucleon  =\cfrac{2.22}{2}=1.11MeV

Question 14

(i) What is meant by Radioactivity?(ii) What is meant by nuclear waste?(iii) Suggest one effective way for the safe disposal of nuclear waste.

Accepted Answer

(i) Radioactivity : It is a nuclear phenomenon. It is the process of spontaneous emission of  \alpha  or  \beta  and  \gamma  radiations from the nuclei of atoms during their decay.(ii) Nuclear Waste : The radioactive material after its use is known as nuclear waste.(iii) The nuclear waste obtained from laboratories, hospitals etc. must be first kept in thick casks and then they must be buried in the specially constructed deep underground stores.

Question 15

Define half-life of a radioactive sample. Which of the following radiations:

α - r a y s, β - r a y s

and

γ - r a y s

(i) are similar to X-rays
(ii) are easily absorbed by matter
(iii) travel with the greatest speed
(iv) are similar in nature of cathode rays ?

Accepted Answer

Half-life: The half-life of a radioactive sample is defined as the time in which the mass of sample is left one half of the origin mass.(i)  \gamma -rays are similar to X-rays(ii)  \alpha -rays are easily absorbed by matter.(iii)  \gamma -rays travel with greatest speed(iv)  \beta -rays are similar to cathode rays.

Question 16

Write three characteristics of nuclear forces.

Accepted Answer

Nuclear force is the force that binds the neutrons and protons in a nucleus together. The following are the few characteristics of nuclear force : -1. It is attractive in nature but with a repulsive core. That is the reason that the nucleus is held together without collapsing in itself2. Nuclear force is identical for all nucleons.3. At a distance less than 0.7 Fermi, this force becomes repulsive. It is one of the most interesting properties of nuclear force, as this repulsive component of the force is what decides the size of the nucleus.

Question 17

State the law of radioactive decay. Plot s graph showing the number (N) of undebased nuclei as a function of time (t) for a given radioactive sample having half-life

T_{1 / 2}

. Depict in the plot the number of undecayed nuclei at
(i)

t = 3 T 1 / 2

and
(ii)

t = 5 T 1 / 2 .

Accepted Answer

Law of radioactive decay : At any instant, the rate of radioactive disintegration is directly proportional to the number of nuclei of the radioactive element present at that instant. where &#955; is the constant of proportionality called radioactive decay constant.Number of undecayed nuclei at t = 3T_{1/2}  is  \dfrac{N_0}{8}  and at  t = 5T_{1/2},  it is  \dfrac{N_0}{32}.

Question 18

a) Three types of radioactive decay occur in nature. Briefly describe them.b) State the law of radio active decay.

Accepted Answer

A) Three types of radio active decay occur in nature: 1. Alpha decay is a radioactive process in which a particle with two neutrons and two protons is ejected from the nucleus of a radioactive atom. The particle is identical to the nucleus of a helium atom. Because alpha particles contain two protons, they have a positive charge of two.2. Beta decay is a radioactive process in which an electron is emitted from the nucleus of a radioactive atom, along with an unusual particle called an antineutrino. The neutrino is an almost massless particle that carries away some of the energy from the decay process. Because this electron is from the nucleus of the atom, it is called a beta particle to distinguish it from the electrons which orbit the atom.3. Gamma decay, type of radioactivity in which some unstable atomic nuclei dissipate excess energy by a spontaneous electromagnetic process. In the most common form of gamma decay, known as gamma emission, gamma rays (photons, or packets of electromagnetic energy, of extremely short wavelength) are radiated. B) According to law of radioactive decay:Number of nuclei decay in a certain time is directly proportional to the total number of nuclei present initially. And number of nuclei decay is directly proportional to time interval. Combining two we get:  \frac{ 	riangle N}{N} = - \lambda 	riangle t Where N = Number of nuclei initially present,  	riangle N  = Number of nuclei decay and  \lambda = Decay constant. After integration we get,  \frac{ 	riangle N}{N_0} = e^{- \lambda t}

Question 19

A nuclear reaction is given as p+ \ ^{15}N\rightarrow  \ _Z^AX+n a) Find  A, Z  and identify the nucleus  X .b) Find the  Q  value of the reaction.c) If the proton were to collide with the  ^{15}N  at rest, find the minimum  KE  needed by the proton to initiate the above reaction

Accepted Answer

a) The nuclear reaction will be  _1^1p+ \ _7^{15}N=\ _Z^AX+ \ _0^1n Now,  1+15=A+1 \Rightarrow A=15  and  1+7=Z+0 \Rightarrow Z=8 The nucleus would be isotope of oxygen i.e  _8^{15}O b) The nuclear reaction will be  _1^1p+\ _7^{15}N=\ _Z^AX+\ _0^1n Now,  1+15=A+1 \Rightarrow A=15  and  1+7=Z+0 \Rightarrow Z=8 The nucleus would be isotope of oxygen i.e  _8^{15}O The atomic mass of proton  =1.00727  amu ; atomic mass of neutron  =1.00866  amu ; atomic mass of  ^{15}O=15.9994  amu and atomic mass of  ^{15}N=15.000108  amuSum of masses of reactants  =1.00727+15.000108=16.0074  amuSum of masses of products  =15.9994+1.00866=17.0081  amuThe Q-value  = Mass difference  \Delta M=17.0081-16.0074=1.0007   amu =1.0007	imes 931 MeV=931.65 MeV c) K.E. of proton to initiate the reaction is equal to Q value of reaction  =	riangle m c^2 	riangle m={ m }_{ (p) }+{ m }_{ (N) }-{ m }_{ (X) }-{ m }_{ (n) }\ =1.007825+15.000-\ ^{ 15 }{ O }-n\ =1.007825+15-15.0031-1.008665\ =-0.00394\  The  K.E needed by the proton to initiate the reaction  Q=-0.00394	imes 931MeV\ =-3.66MeV Negative sign signifies that this energy is supplied to initiate the reaction.

Question 20

(a) Plot a graph showing the variation of undecayed nuclei  N  versus time  t . From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.(b) The air in some caves includes a significant amount of radon gas, which can lead to lung cancer if breathed over a prolonged time. In British caves, the air in the cave with the greatest amount of the gas has an activity per volume of  1.55 	imes 10^5 Bq/m^3  . Suppose that you spend two full days exploring (and sleeping in) that cave. Approximately how many  ^{222}Rn  atoms would you take in and out of your lungs during your two-day stay? The radionuclide  ^{222}Rn  in radon gas has a half-life of 3.82 days. You need to estimate your lung capacity and average breathing rate.

Accepted Answer

(a) Law of Radioactivity defines that the number of Nuclei undergoing number of Nuclei present in the sample at that Instant.Since from the graph; we have      N=N_{ - }e^{-\lambda t }     ....1where  \lambda  =Disintegration constantFor  	herefore   For  T_{1/2}  is the time at  N-\frac{1}{2}N_D \Rightarrow   t=T_{1/2} \frac{N_0}{2} = N_0e^{-\lambda\frac{T}{2}} \Rightarrow   \boxed { T_{ 1/2 }=\frac { Ln2 }{ \lambda  } =\frac { 0.693 }{ \lambda   }  }  And for Mean life -we have to sum it over the whole Range for  N(t)=N_0e^{-\lambda t} for number of nuclei which decay in time t to t  t  	riangle ; N(t)	riangle t=\lambda N_{ 0 }e^{ -\lambda  \,  t}	riangle t  For Integration it over the Range  T=0\, to\,\infty 	au = \frac { \lambda N_{ 0 }\iint _{ 0 }^{ \infty  }{ te^{ -\lambda t }dt }  }{ N_ 0 } =\lambda \int ^{ \infty  }_0{ t } e^{ \lambda t }dt \boxed { 	au =\frac { 1 }{ \lambda }  }  Mean-life(b) The equation for the activity is given by: R=\lambda N Here, R is the activity, N is the number of nuclei and  \lambda  is the decay constant. The equation for the decay constant is given by,  \lambda=\dfrac{ln\,2}{T_{1/2}} Here,  T_{1/2}  is the half - lifeThus,  R=\dfrac{ln\,2}{T_{1/2}}N N=\dfrac{RT_{1/2}}{ln\,2} Dividing by volume,  \dfrac{N}{V}=\dfrac{R}{V}\dfrac{T_{1/2}}{ln\,2} Substituting the value, =\dfrac{N}{V}=(1.55	imes 10^5Bq/m^3)\dfrac{3.82	imes 24	imes 3600s}{ln\,2}              =7.38	imes 10^{10}     atoms/m^3 Thus, these are  7.38	imes 10^{10}  atoms  /m^3Rn  atms per unit volume in the cave.