In the following graph binding energy per nucleon is plotted against mass number (A). Three elements $A_{1}, A_{2}, A_{3}$ are located in the graph. Select the false statement about the graph.

Element

A_{2}

is more stable than

A_{1}

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Element

A_{3}

is less stable than

A_{2}

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Element

A_{1}

is more stable than both

A_{2}

and

A_{3}

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A_{2}

is metallic element

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Solution

The correct option is A Element $A_{2}$ is more stable than $A_{1}$

The binding energy per nucleon and the mass number is plotted in the figure. We can see that the curve rises steeply initially and then gradually reaches maximum of .The curve now drops very slowly to somewhere around the highest known mass number. From this graph we can say that nuclei of intermediate mass are more stable since high amount of energy is required to separate these nucleons. This also means that a large amount of energy will be liberated if heavier nuclei can be divided into lighter ones in nuclear fission.
Hence option A is correct.

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Similar questions

\frac{a_{1}}{a_{1} + 1 -} \frac{a_{2}}{a_{2} + 1 -} \frac{a_{3}}{a_{3} + 1 -} \dots, a_{1}, a_{2}, a_{3}, \dots

a_{1}, a_{2}, a_{3} . . . a_{n}

a_{3}

a_{1}, a_{2}, a_{3}, . . . a_{n}

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

A = a_{1}, a_{2}, a_{3} . . ., a_{n}

a_{3}

=

If A₁, A₂, A₃,..., A_n are n numbers between a and b, such that a, A₁, A₂, A₃,..., A_n, b are in A.P., then

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