
Radius of nucleus varies as $R = {R_0}{\left( A \right)^{\dfrac{1}{3}}}$ , where ${R_0} = 1.3$ Fermi. What is the volume of $B{e^8}$ nucleus (approx.) [ A=atomic mass]?
A. $7 \times {10^{ - 38}}cc$
B. $7 \times {10^{ - 29}}cc$
C. $7 \times {10^{ - 45}}cc$
D. None of these
Answer
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Hint: We need to substitute the value of the atomic mass in the given expression to calculate the radius of the $B{e^8}$ nucleus. We need to calculate the volume of the $B{e^8}$ nucleus. The volume of an atom is given as $\dfrac{{4\pi }}{3}{R^3}$ Here $R$ is the radius of the nucleus of the atom. The given unit of radius is in Fermi and the options are in cubic centimetre.
Complete step by step answer:
We are given with the expression to calculate the radius of the nucleus of an atom. We are given with a $B{e^8}$ atom. The atomic mass of an atom is given as $8$ .
The radius of the nucleus of the $B{e^8}$ atom will be given as:
$R = {R_0}{\left( 8 \right)^{\dfrac{1}{3}}}$
$ \Rightarrow R = {R_0} \times 2$
We are given that ${R_0} = 1.3$ Fermi, therefore the radius of the $B{e^8}$ atom’s nucleus will be:
$R = 1.3 \times 2$ Fermi
$ \Rightarrow R = 2.6$ Fermi
The relation between Fermi and centimetre is $1\,Fermi = {10^{ - 13}}cm$
$ \Rightarrow R = 2.6 \times {10^{ - 13}}cm$
Therefore, the radius of the nucleus of $B{e^8}$ atom is $R = 2.6 \times {10^{ - 13}}cm$
The volume of the nucleus will be given as:
$V = \dfrac{{4\pi }}{3}{R^3}$
Here, $V$ is the volume.
Substituting the value of radius of the nucleus of $B{e^8}$ atom, we get
$V = \dfrac{{4\pi }}{3}{\left( {2.6 \times {{10}^{ - 13}}} \right)^3}$
$ \Rightarrow V = 4.18 \times {\left( {2.6 \times {{10}^{ - 13}}} \right)^3}$
$ \Rightarrow V = 7.36 \times {10^{ - 38}}cc$
$ \Rightarrow V \approx 7 \times {10^{ - 38}}cc$
This is the approximate volume of the $B{e^8}$ nucleus.
So, the correct answer is “Option A”.
Additional Information:
Chronic beryllium disease also known as CBD is responsible for causing scarring of the lung tissue. It occurs when a person inhales dust or fumes of beryllium which is a naturally occurring lightweight material.
Note:
The given unit of radius was in Fermi, converting it to centimeters.
The given expression for radius is for the radius of the nucleus and not the radius of the atom.
Radius of the nucleus is proportional to the cube root of its atomic mass.
All the mass of an atom is considered to be present in the nucleus.
Complete step by step answer:
We are given with the expression to calculate the radius of the nucleus of an atom. We are given with a $B{e^8}$ atom. The atomic mass of an atom is given as $8$ .
The radius of the nucleus of the $B{e^8}$ atom will be given as:
$R = {R_0}{\left( 8 \right)^{\dfrac{1}{3}}}$
$ \Rightarrow R = {R_0} \times 2$
We are given that ${R_0} = 1.3$ Fermi, therefore the radius of the $B{e^8}$ atom’s nucleus will be:
$R = 1.3 \times 2$ Fermi
$ \Rightarrow R = 2.6$ Fermi
The relation between Fermi and centimetre is $1\,Fermi = {10^{ - 13}}cm$
$ \Rightarrow R = 2.6 \times {10^{ - 13}}cm$
Therefore, the radius of the nucleus of $B{e^8}$ atom is $R = 2.6 \times {10^{ - 13}}cm$
The volume of the nucleus will be given as:
$V = \dfrac{{4\pi }}{3}{R^3}$
Here, $V$ is the volume.
Substituting the value of radius of the nucleus of $B{e^8}$ atom, we get
$V = \dfrac{{4\pi }}{3}{\left( {2.6 \times {{10}^{ - 13}}} \right)^3}$
$ \Rightarrow V = 4.18 \times {\left( {2.6 \times {{10}^{ - 13}}} \right)^3}$
$ \Rightarrow V = 7.36 \times {10^{ - 38}}cc$
$ \Rightarrow V \approx 7 \times {10^{ - 38}}cc$
This is the approximate volume of the $B{e^8}$ nucleus.
So, the correct answer is “Option A”.
Additional Information:
Chronic beryllium disease also known as CBD is responsible for causing scarring of the lung tissue. It occurs when a person inhales dust or fumes of beryllium which is a naturally occurring lightweight material.
Note:
The given unit of radius was in Fermi, converting it to centimeters.
The given expression for radius is for the radius of the nucleus and not the radius of the atom.
Radius of the nucleus is proportional to the cube root of its atomic mass.
All the mass of an atom is considered to be present in the nucleus.
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