NCERT MOST IMPORTANT QUESTIONS CLASS – 12 | PHYSICS IMPORTANT QUESTIONS | CHAPTER – 13 | NUCLEI | EDUGROWN |

In This Post we are  providing Chapter- 13 NUCLEI NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON NUCLEI

Question 1.
Calculate the binding energy per nucleon of Fe⁵⁶₂₆ Given m_Fe = 55.934939 u, m_n = 1.008665 u and m_p = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z m_p + (A – Z)m_n – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 10²³ nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10^-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 10⁹ years. Calculate the activity of 1 g sample of ₉₂²³⁸U.
Answer:
Given T = 4.5 × 10⁹ years.
Number of nuclei of U in 1 g
= N = 6.023×1023238 = 2.5 × 10²¹

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day^-1
or
T_1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = N_oe^-λt
or
N = N_o e^{-0 3456 × 4}
or
N = N₀ e^{-1 3824} = N_o × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/N_o = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day^-1

Question 7.
A radioactive substance decays to 1/32^th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = N_o / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T₁/2 = 30 s, N = 3N_o / 4, λ = ?, t = ?
(i) Decay constant
λ = 0.693T1/2=0.69330 = 0.0231 s^-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = N_oe^-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given N_ox = N_oY, T_X = 1 h, T_Y = 2 h, therefore
λXλY=21 = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are R_x and R_y respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 10³² kg and it generates energy at the rate of 5 × 10³⁰ watt. How long will it take to convert all the helium to carbon at this rate?

Answer:

As 4 × 10^-3 kg of He consists of 6.023 × 10²³ He nuclei so 5 × 10³² kg He will contain
6.023×1023×5×10324×10−3 = 7.5 × 10⁵⁸ nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10^-13 J of energy
So all nuclei in the star will produce
E = 7.27×1.6×10−133 × 7.5 × 10⁵⁸
= 2.9 × 10⁴⁶ J

As power generated is P = 5 × 10³⁰ W, therefore time taken to convert all He nuclei into carbon is
t = EP=2.9×10465×1030 = 5.84 × 10¹⁵ s
or
1.85 × 10⁸ years