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A match box of $5cm \times 5cm \times 1cm$ dimensions is filled with nuclear matter. Its weight is in the order of
(A) $10g$
(B) ${10^8}g$
(C) ${10^{12}}g$
(D) ${10^{15}}g$

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Last updated date: 13th Jun 2024
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Answer
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Hint:A nuclear matter is a system which has two fermions namely protons and neutrons. These protons and neutrons interact with each other only due to nuclear forces. So, overall, we have a nucleus and this nucleus must have some volume, find this volume. After finding the volume of the nucleus, find how many such nuclei can fit in the match box. After knowing the number of nuclei, find the mass of the total matter present.

Complete step by step solution:
The average radius of a nucleus is given by the formula $R = {R_0}{A^{\dfrac{1}{3}}}$, where the value of ${R_0}$ is equal to $1.1 \times {10^{ - 15}}$$m$ and $A$ is the mass number.

Assume the nucleus to be spherical in shape, the volume of the nucleus will be given as $V =
\dfrac{4}{3}\pi {R^3}$. Let us substitute the value of $R$ in terms of ${R_0}$ and $A$, we will get the volume of nucleus as
\[
V = \dfrac{4}{3}\pi {\left( {{R_0}{A^{\dfrac{1}{3}}}} \right)^3} \\
V = \dfrac{4}{3}\pi {R_0}^3A \\
\]

So, this volume corresponds to one nucleus. Now, if this sample of nucleus is filled in a matchbox of
$5cm \times 5cm \times 1cm$ dimensions, more number of nuclei are required, Let the number of nuclei required be $n$. The total volume occupied by these nuclei must be equal to the number of nuclei present multiplied by the volume of one nucleus, mathematically, we will have total volume occupied equal to $\dfrac{4}{3}\pi {R_0}^3An$. The obtained value of the total volume must be equal to the volume of the matchbox, that is \[25c{m^3} = 25 \times {10^{ - 6}}{m^3}\]. Therefore,

we have,
$
\dfrac{4}{3}\pi {R_0}^3An = 25 \times {10^{ - 6}} \\
\Rightarrow n = 25 \times {10^{ - 6}} \times \dfrac{3}{{4\pi }} \times \dfrac{1}{{{R_0}^3A}} \\
\Rightarrow n = 25 \times {10^{ - 6}} \times \dfrac{3}{{4\pi }} \times \dfrac{1}{{{{\left( {1.1 \times {{10}^{ -
15}}} \right)}^3}A}} \\
\Rightarrow n \approx \dfrac{{4.5 \times {{10}^{39}}}}{A} \\
$

So, there are a total number of $\dfrac{{4.5 \times {{10}^{39}}}}{A}$ nuclei.
Now, the mass of a nucleus is given as $M = mA$, where $m$ is the mass proton and $A$ is the mass number.

So, the mass of the nuclear matter will be the number of nuclei multiplied by its mass. So,
${M_{total}}$ will be given as,
$
{M_{total}} = \dfrac{{4.5 \times {{10}^{39}}}}{A} \times mA \\
\Rightarrow {M_{total}} = 4.5 \times {10^{39}} \times 1.66 \times {10^{ - 27}} \\
\Rightarrow {M_{total}} = 6.972 \times {10^{12}}kg \\
\Rightarrow {M_{total}} = 6.972 \times {10^{15}}g \\
$

Therefore, the mass of the nuclear matter in a matchbox of $5cm \times 5cm \times 1cm$
dimensions is in the order of ${10^{15}}g$.

Thus, the correct Option is (D).

Note: Here, we have considered a general case in which we have taken the element of the nuclear matter to have a mass number of $A$. Also remember the method through which we found the mass of the nuclear matter, we found the number of nuclei present in the matchbox using the volume analysis and then found the total mass of the nuclear matter.

Keep in mind the formulae used in this solution. The most important thing to notice is we have found the mass of the nuclear matter and not the weight. But, the most appropriate option to go with was option D.

Actually, the order of the weight should be ${10^{16}}g$, assuming that the $'g'$ mentioned in the options represents the unit gram. If it represents the gravitational constant, then there is nothing to be worried about, option D is the correct answer.