Q: List- I shows different radioactive decay processes and List- II provides possible emitted particles. Match each entry in List- I with an appropriate entry from List- II , and choose the correct optio

In α decay mass number decreases by 4 unit and atomic number decreases by 2 unit. In β - decay mass number does not change but atomic number increases by 1 unit. In β + decay mass number does not change but atomic number decreases by 1 unit. So for P 238 92 U → 234 91 P a . Number of alpha particle = 238 - 234 4 = 1 . Due to alpha particle, atomic number should decrease by 2 but it has decreased by 1 only, therefore&nbsp

Q: In a radioactive decay chain reaction, 90 230 Th nucleus decays into Po 84 214 nucleus. The ratio of the number of α to number of β - particles emitted in this process is _

Let number of α particles are n α and β particles are n β . Given: 90 Th 230 → 84 Po 214 + n α α + n β β . Change in mass number is = 230 - 214 = 4 n α ⇒ n α = 4 Number of proton, n β = 84 - 90 - 2 n a ⇒ n β = 2 So, the ratio, n α n β = 2

Q: The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction N 7 16 + He 2 4 → H 1 1 + O 8 19 in a laboratory frame is n (in MeV ). Assume that N 7 16 is at rest in

Given here: N 7 16 + He 2 4 → H 1 1 + O 8 19 Energy absorbed in the reaction is Q = m N + m He - m H - m O × c 2 = 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 × 930 MeV = - 1 . 86 MeV = 1 . 86 MeV Let v is velocity of alpha particle. Now, the maximum loss of kinetic energy = 1 2 × m × 4 m 5 m × v 2 ⇒ 1 2 m v 2 = 5 4 × Q = 5 4 × 1 . 86 MeV = 2 . 325 MeV ∴ n = 2 . 33

Q: A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60 % and beta-decay with probability of 40 % . Initially, the number of&n

Given, T 1 2 = 20 min N 0 = 1000 ∵ 1 hour = 60 min = 3 T 1 2 ∴ Number of nuclei remain after 1 hour, N = N o 2 3 = 1000 8 ⇒ N = 125 It means 875 nuclei has decayed, Now, Number of α decayed Nucleus = 875 × 60 100 = 525

Question 1

A nuclear reactor starts producing a radioactive nuclide X from t = 0, at a constant rate of per second. Each decay of X produces energy E_0, which is utilized to heat a liquid of mass m and specific

Accepted Answer

Let N be the number of active nuclei at time t. The rate of change of the number of nuclei is given by dN dt = - N Integrating with the initial condition N = 0 at t = 0: _ 0 ^ N dN - N = _ 0 ^ t dt - 1 ( - N ) = t N = (1 - e^ - t ) The rate of decay of the nuclide at time t is A = N = (1 - e^ - t ) Since each decay produces energy E_0, the rate of energy production is dE dt = A E_0 = E_0 (1 - e^ - t ) This energy is used to heat the liquid. The rate of heat absorption is dQ dt = m s dT dt Equati

Question 2

In a radioactive decay process, the activity is defined as  A = - d N d t , where N ( t ) is the number of radioactive nuclei at time t . Two radioactive sources, S 1  and S 2 &nbs

Accepted Answer

Activity for both the cases can be written as,  A 1 = A 0 e - &#955; 1 t 1 also  A 2 = A 0 e - &#955; 2 t 2 . Half-life for the reaction can be written as,  t 1 2 1 = ln 2 &#955; 1  and hence at  t 1 = 3 ln 2 &#955; 1 . A 1 = A 0 e - &#955; 1 3 ln 2 &#955; 1 &#8658; A 1 = A 0 e - 3 ln 2 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . &#160; i Similarly, for the second case, we can write t 2 = 7 ln 2 &#

Question 3

List- I shows different radioactive decay processes and List- II provides possible emitted particles. Match each entry in List- I with an appropriate entry from List- II , and choose the correct optio

Accepted Answer

In &#945;  decay mass number decreases by 4 unit and atomic number decreases by 2 unit. In &#946; -  decay mass number does not change but atomic number increases by 1 unit. In &#946; +  decay mass number does not change but atomic number decreases by 1 unit. So for P &#160; 238 92 U &#8594; 234 91 P a . Number of alpha particle  = 238 - 234 4 = 1 . Due to alpha particle, atomic number should decrease by  2  but it has decreased by  1  only, therefore&nbsp

Question 4

In a radioactive decay chain reaction, 90 230 Th  nucleus decays into Po 84 214  nucleus. The ratio of the number of &#945; to number of &#946; -  particles emitted in this process is _

Accepted Answer

Let number of  &#945;  particles are  n &#945;  and  &#946;  particles are  n &#946; . Given:  90 Th 230 &#8594; 84 Po 214 + n &#945; &#945; + n &#946; &#946; . Change in mass number is  = 230 - 214 = 4 n &#945; &#8658; n &#945; = 4 Number of proton,  n &#946; = 84 - 90 - 2 n a &#8658; n &#946; = 2 So, the ratio,  n &#945; n &#946; = 2

Question 5

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction N 7 16 + He 2 4 &#8594; H 1 1 + O 8 19 in a laboratory frame is n  (in MeV ). Assume that N 7 16 is at rest in

Accepted Answer

Given here:  N 7 16 + He 2 4 &#8594; H 1 1 + O 8 19 Energy absorbed in the reaction is  Q = m N + m He - m H - m O &#215; c 2 = 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 &#215; 930 &#160; MeV = - 1 . 86 &#160; MeV = 1 . 86 &#160; MeV   Let  v  is velocity of alpha particle. Now, the maximum loss of kinetic energy =   1 2 &#215; m &#215; 4 m 5 m &#215; v 2 &#8658; 1 2 m v 2 = 5 4 &#215; Q = 5 4 &#215; 1 . 86 &#160; MeV = 2 . 325 &#160; MeV &#8756; &#160; n = 2 . 33

Question 6

The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton

Accepted Answer

Total binding energy (without considering repulsions), E b = Z m p + A - Z m n - m x c 2 Where, X Z A is the nuclei under consideration. Now, considering repulsion : Number of proton pairs  = C 2 Z &#8658; Thus repulsion energy  &#8733; Z Z - 1 2 &#215; 1 4 &#960; &#1013; 0 e 2 R Where R  is the radius of the nucleus &#8658; E b p - E b n &#8733; Z Z - 1 &#160;&#160; &#8756; there will be no repulsion term for neutrons. Also, since  R = R 0 A 1 3 &#8658; &#160; E b p - E b n

Question 7

A heavy nucleus  N , at rest, undergoes fission  N &#8594; P + Q ,  where  P  and  Q  are two lighter nuclei. Let  &#948; = M N - M P - M Q , where  M P ,

Accepted Answer

Since,  F e x t = 0 So, According to conservation of linear momentum, p total &#160; initial = p total &#160; final = 0 That's why momentum of both lighter nuclei will be equal and opposite to each other as we can see in above diagram. Now here some energy released due to mass defect which can be written as, Energy released  = &#916; m c 2 = &#948; c 2 Where,  &#916; m = &#948; = mass defect and,  &#948; = M N - M P - M Q Now this energy released

Question 8

A heavy nucleus  Q  of half-life  20  minutes undergoes alpha-decay with probability of  60 %   and beta-decay with probability of  40 % . Initially, the number of&n

Accepted Answer

Given, T 1 2 = 20 &#160; min N 0 = 1000 &#8757; &#160; &#160; 1 &#160; hour = 60 &#160; min = 3 T 1 2 &#8756;  Number of nuclei remain after  1  hour, N = N o 2 3 = 1000 8 &#8658; N = 125 It means  875  nuclei has decayed, Now, Number of  &#945;  decayed Nucleus = 875 &#215; 60 100 = 525

Nuclear Physics — Previous Year Questions & Solutions