Questions in Nuclear Physics at the level expected of you are simple and interesting. You should be careful not to omit questions from this section. Consider the following MCQ which appeared in Kerala Medical Entrance 2000 test paper:

A = A

[The well known radioactive decay law is, N = N

Therefore, we have, 6.25 = 100/2

Let us discuss another MCQ:

Mean life, T = 1/λ = T

Let us consider one more question involving radio activity:

You can work this out even if you don’t have a calculator or logarithm tables since you definitely remember that log 2 is 0.3010. Therfore, n = 1/0.3010 = 3.32.

So you require 3.32 half lives or, 3.32×3.02 days = 10 days for the sample to decay to 10%.

Now consider the following question:

The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR

You can easily show that the

Mass of nucleus = 1.67×10

Volume of nucleus = (4/3)πR

But, R = 1.1×10

d = (1.67×10

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

**A radioactive isotope has a half life T years. The time after which its activity is reduced to 6.25% of its original activity is**

(a) 2T years (b) 4T years (c) 6T years (d) 8T years (e) 16T years

Since the activity of a sample is directly proportional to the number of nuclei present at the instant, we can express the activity ‘A’ after ‘n’ half lives in terms of the initial activity ‘A(a) 2T years (b) 4T years (c) 6T years (d) 8T years (e) 16T years

_{0}’ as,A = A

_{0}/2^{n.}[The well known radioactive decay law is, N = N

_{0}e^{-λt }where N_{0}is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant. This equation, modified in terms of half life can be written as N = N_{0}/2^{n}where N is number of nuclei remaining undecayed after ‘n’ half life periods. Since the activity, A = dN/dt = -λN, it follows that the activity also can be expressed in the same manner as we express N. Therfore, A = A_{0}/2^{n}].Therefore, we have, 6.25 = 100/2

^{n}, taking initial activity as 100. This yields n = 4, which means that 4 half lives have been elapsed to reduce the activity to 6.25% of the initial activity. The time required is therefore 4T, given in option (b).Let us discuss another MCQ:

**Out of 10 mg of a radio active sample, 1.25 mg remains undecayed after 6 hours. The mean life of the sample in hours is**

(a) 0.693 (b) 2/0.693 (c) 4/0.693 (d) 0.693/4 (e) 0.693/2

We have, N = N(a) 0.693 (b) 2/0.693 (c) 4/0.693 (d) 0.693/4 (e) 0.693/2

_{0}/2^{n}so that 1.25 = 10/2^{n}, from which 2^{n}= 8 and n = 3. So, 3 half life periods is 6 hours so that the half life of the sample is 2 hours.Mean life, T = 1/λ = T

_{half }/0.693 = 2/0.693 [Option (b)].Let us consider one more question involving radio activity:

**The half life of a radioactive sample is 3.02 days. After how many days 10% of the sample will remain undecayed?**

(a) 10 (b) 12.5 (c) 15 (d) 20 (e) 25

Using the relation, N = N(a) 10 (b) 12.5 (c) 15 (d) 20 (e) 25

_{0}/2n where ‘n’ is the number of half life periods in which the sample decays from N_{0}to N, we have, 10 = 100/2^{n}. From this, 2^{n}= 10. Taking logarithms, n log 2 = log 10.You can work this out even if you don’t have a calculator or logarithm tables since you definitely remember that log 2 is 0.3010. Therfore, n = 1/0.3010 = 3.32.

So you require 3.32 half lives or, 3.32×3.02 days = 10 days for the sample to decay to 10%.

Now consider the following question:

**The density ‘d’ of nuclear matter varies with nucleon number ‘A’ as**

(a) d α A

The correct option is (e).(a) d α A

^{-1}(b) d α A^{-2}(c) d α A (d) da A^{2}(e) d α A^{0}The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR

^{3}where R is the nuclear radius. But, R = 1.1×10^{-15}A^{⅓}. So, the volume of the nucleus also is directly proportional to the nucleon number A. Since density is the ratio of mass to volume, it follows that the*** * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ****density of nuclear matter is independent of the nucleon number A.*

You can easily show that the

**as follows:***density of nuclear matter is of the order of 10*^{17}kg/m^{3}Mass of nucleus = 1.67×10

^{-27}×A since the mass of a nucleon is approximately 1.67×10^{-27}kgVolume of nucleus = (4/3)πR

^{3}where R is the nuclear radius.But, R = 1.1×10

^{-15}A^{⅓}metre so that density of nuclear matter,d = (1.67×10

^{-27}×A) /[(4/3)π×(1.1×10^{-15}× A^{⅓})^{3}. This works out to approximately 3×10^{17}kg/m^{3}.**The following simple MCQ appeared in AIEEE 2004 question paper:**

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be

(a) 2

Since the momenta of the parts have to be equal in magnitude in accordance with the law of consevation of momentum, we have, mA nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be

(a) 2

^{⅓}:1 (b) 1:2^{⅓}(c) 3^{½}:1 (d) 1:3^{½}_{1}v_{1}= m_{2}v_{2}so that m_{1}/m_{2}= v_{2}/v_{1}=1/2. The masses being directly proportional to the volumes, and the volumes being directly proportional to the cube of the radii, the radii (sizes) are directly proportional to the cube root of the masses. So, the answer is 1:2^{⅓}.* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

**Suppose a radioactve mother nucleus emits a β-particle. Are the mother and daughter nuclei isotones or isobars?**When a nucleus emits a β-particle, a neutron in the nucleus becomes a proton and hence the neutron number is changed. So the mother and daughter are not isotones. The mass number is not changed but the proton number (atomic number) is changed. So they are isobars.

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

Take note of the following question on calculation of the activity of a sample:

Number of nuclei (or, atoms) in 2 gram of radium is (2/226)×N

Activity = λN = (0.693/T

Note that we have converted the half life in years into seconds. [ 1 year = 325.25×24×60×60 seconds = 3.16×10

The activity works out to 7.2×10

Now consider the following MCQ which involves Einstein’s mass- energy relation:

[We have converted megawatt into watts and hour into seconds].

Since this is equal to mc

m = (3.6×10

Take note of the following question on calculation of the activity of a sample:

**Half life of radium (**

Note that becquerel (Bq) is the unit of radioactivity in SI and is equal to one disintegration per second._{88}Ra^{226}) is 1620 years. What is the activity of 2gram of radium? (a) 3.6×10^{10}Bq (b) 7.2×10^{10}Bq (c) 1.44×10^{11}Bq (d) 2.88×10^{11}Bq (e) 5.76×10^{11}BqNumber of nuclei (or, atoms) in 2 gram of radium is (2/226)×N

_{A}= (2/226)×6.025×10^{23}= 5.33×10^{21}. [N_{A}is the Avogadro number, 6.025×10^{23}].Activity = λN = (0.693/T

_{half})×N = [0.693/(1620×3.16×10^{7})] × (5.33×10^{21}).Note that we have converted the half life in years into seconds. [ 1 year = 325.25×24×60×60 seconds = 3.16×10

^{7}s].The activity works out to 7.2×10

^{10}becquerel.Now consider the following MCQ which involves Einstein’s mass- energy relation:

**How much mass has to be converted into energy to produce electric power of 100 MW for one hour? (Assume that the conversion efficiency is 100%).**

(a) 1mg (b) 2 mg (c) 4 mg (d) 20 mg (e) 40mg

At the rate of 100 MW, total energy produced for 1 hour is, P×t = (100×10(a) 1mg (b) 2 mg (c) 4 mg (d) 20 mg (e) 40mg

^{6}) ×(60×60) joule = 3.6×10^{11}J[We have converted megawatt into watts and hour into seconds].

Since this is equal to mc

^{2}in accordance with Einstein’s mass energy relation, we have,m = (3.6×10

^{11})/(3×10^{8})^{2}= 4×10^{-6}kg = 4 mg.Here is a question involving relativistic increase of mass:

**The rest mass of a proton is ‘m**

(a) 3m

The mass of the proton while moving with velocity ‘v’ is given by_{0}’. Its linear momentum when it moves with half the speed of light ‘c’ in free space is(a) 3m

_{0}c/4 (b) m_{0}c/2 (c) m_{0}c (d) 2m_{0}c/√3 (e) m_{0}c/√3m = m

_{0}/√[1- v^{2}/c^{2}] so that when v = c/2, m = m_{0}/√[1- ¼] = 2m_{0}/√3.The momentum of the proton is mc/2 = m

_{0}c/√3.
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