Question

Fission of nuclei is possible because the binding energy per nucleon in them

A. increases with mass number at high mass numbers
B. decreases with mass number at high mass numbers  
C. increases with mass number at low mass numbers
D. decreases with mass number at low mass numbers
Answer :   decreases with mass number at high mass numbers
Solution :
The binding energy per nucleon for the middle nuclides (from $$A = 20$$  to $$A = 56$$  ) is maximum. Hence, these are more stable.
As the mass number increases, the binding energy per nucleon gradually decreases and ultimately binding energy per nucleon of heavy nuclides (such as uranium etc.) is comparatively low. Hence, these nuclides are relatively unstable. So, they can be fissioned easily.

Releted MCQ Question on
Modern Physics >> Atoms or Nuclear Fission and Fusion

Releted Question 1

The equation
$$4_1^1{H^ + } \to _2^4H{e^{2 + }} + 2{e^ - } + 26MeV$$       represents

A. $$\beta $$ -decay
B. $$\gamma $$ -decay
C. fusion
D. fission
Releted Question 2

Fast neutrons can easily be slowed down by

A. the use of lead shielding
B. passing them through water
C. elastic collisions with heavy nuclei
D. applying a strong electric field
Releted Question 3

In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$    the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$    ]

A. $${10^7}K$$
B. $${10^5}K$$
C. $${10^3}K$$
D. $${10^9}K$$
Releted Question 4

The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$  is $$1.1\,MeV$$  and $$7\,MeV$$  respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

A. $$23.6\,MeV$$
B. $$26.9\,MeV$$
C. $$13.9\,MeV$$
D. $$19.2\,MeV$$

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