Model NEET Exam Physics Questions with Answer

Nuclear Fission

Very large nuclei having mass number greater than 230 tend to be unstable and can split into two or more parts. This is called Fission. Fission process is not a spontaneous one. It occurs when a neutron strikes an unstable nucleus.

In this case, the Uranium 235 will split into two nearly equal nuclei Barium-141 and Krypton-92 and release several free neutrons with huge amounts of energy.

₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3 ₀n¹ + Energy

₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹ + Energy

These nuclei are isotopes of more stable elements. If left alone, they decay radioactively by emitting alpha or beta particles. The below figure clearly explain how the chain reaction takes place. Out come neutrons from fission process interact with fissile fuel present in the surrounding. Further this reaction produces totally nine neutrons. This cycle repeats to give a reaction.  Moderator like graphite, heavy water and beryllium are used to control the rate of reaction. If the chain reaction is not controlled, a nuclear explosion will occur.

CALCULATING  ENERGY RELEASED BY FISSION

Calculate the energy released in the following spontaneous fission reaction:

²³⁸U → ⁹⁵Sr + ¹⁴⁰Xe + 3n

given the atomic masses to be m(²³⁸U) = 238.050784 u, m(⁹⁵Sr) = 94.919388 u, m(¹⁴⁰Xe) = 139.921610 u, and m(n) =1.008665 u.

Solution

 The products have a total mass of

m_products = 94.919388 u+139.921610 u+3(1.008665 u) = 237.866993 u

The mass lost is the mass of ²³⁸U minus m_products, or

Δm = 238.050784 u − 237.8669933 u = 0.183791 u

so the energy released is E = (Δm) c² = 171.2 MeV

Energy released by 1 Kg of uranium:

Number of atoms in 1 kg of uranium = (6.023 × 10²⁶)/235

Energy released in one fission = 200 MeV

Energy produced by 1 kg of uranium during fission

E = (6.023 × 10²⁶ × 200)/235 = 5.128 × 10²⁶ MeV

E = (5.128 × 10²⁶) × (1.6×10^-13) J      [1 MeV = 1.6×10^-13 J]

E = {(5.128 × 10²⁶) × (1.6×10^-13)}/{3.6× 10⁶} k Wh

E = 2.26 × 10⁷ k Wh                                    [1k Wh = 3.6×10⁶ J]

Due to this reason, nuclear energy is being used for the generation of electricity. 

Most of the energy that is released during fission goes into the K.E of the fission fragments. The emitted neutrons, Beta and gamma rays and neutrinos carry off perhaps 20% of the total energy.

Read it carefully and try to solve simple problems Click Here

Nuclear Transmutation: Objective Type Questions

1) Which reaction converts an atom of one element to an atom of another element?

a) combustion                                    b) polymerisation          

c) transmutation                                d) Saponification

2) Radioactive Cobalt-60 is used in radiation therapy treatment. Cobalt-60 undergoes beta decay. This type of nuclear reaction is called __________

a) artificial transmutation        b) natural transmutation

c) nuclear fission                     d) nuclear fusion

3) Which is most effective for artificial transmutation?

a) Proton                         b) Deuteron           

c) Helium nuclei             d) Neutron

Answer: Neutrons possess no charge at all and are considered to be the best projectiles.

4) For artificial transmutation of nuclei, the least effective one is

a) Proton                          b) Deuteron           

c) Alpha particles           d) Neutron

Answer: Due to heavy mass, alpha particles cannot easily pass through matter.

5) The given nuclear reaction is X + ₀n¹ → ₃Li⁷ + ₂He⁴. The particle represented by X is __________

a) ₄Li⁹          b) ₅B¹⁰         c) ₅Be¹⁰        d) ₆C¹⁰

Answer: ₅B¹⁰+ ₀n¹ → ₃Li⁷ + ₂He⁴   Here Z and A are balanced.

6) ₁₃Al²⁸ is transferred into ₁₅P³¹ and neutron when it is bombarded with suitable projectile. The projectile used is _________

a) Proton                         b) Deuteron           

c) Alpha particle            d) Neutron

Answer: ₁₃Al²⁸ + ₂He⁴   → ₁₅P³¹ + ₀n¹ Here Z and A are balanced.

7) Artificial elements have been prepared by bombardment reactions in high energy accelerators. Find the mass number of element X produced in the given reaction. 

₉₅Cf²⁴⁹ + ₇N¹⁵   → ₁₀₅X + 4 ₀n¹

a) 261         b) 260         c) 264          d) 257

Answer: A= (249+15-4) = 260

8) (JEE 2011 Question) Bombardment of aluminium by alpha particle leads to its artificial disintegration in two ways, (i) and (ii) as shown below. Products X, Y and Z respectively are: 

a) Proton, Neutron, Positron           b) Neutron, Positron, Proton

c) Proton, Positron, Neutron              d) Positron, Proton, Neutron

 Answer:     For X, ₁₃Al²⁷ + ₂He⁴   → ₁₄Si³⁰ + ₁H¹

For Y, ₁₃Al²⁷ + ₂He⁴   → ₁₅P³⁰ + ₀n¹

                   For Z, ₁₅P³⁰ → ₁₄Si³⁰ + ₁e⁰

9) (JEE 2013 Question) 

In the nuclear transmutation, ₄Be⁹ + X → ₄Be⁸ + Y. 

(X,Y) is (are)

a) (γ,n)        b) (P,D)       c) (n,D)        d) (γ,D)

Answer: A and B

₄Be⁹ + γ → ₄Be⁸ + ₀n¹

₄Be⁹ + ₁H¹ → ₅B¹⁰ → ₄Be⁸ + ₁H²

10) (JEE 2012 Question) The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. The group to which, element X belongs in the periodic table is ______

₂₉Cu⁶³ + ₁H¹ → 6 ₀n¹ + ₂He⁴ + 2 ₁H¹ + X

Answer: Group 8 element : ₂₆Fe⁵²

Element X : Atomic number Z = (29+1)-(0+2+2)=26

                   Mass number A = (63+1)-(6+4+2)=52

Hence X is ₂₆Fe⁵². It belongs to group 8.