
Two spherical nuclei have mass number 216 and 64 with their radii ${R_1}$ and ${R_2}$ respectively.
The ratio, $\dfrac{{{R_1}}}{{{R_2}}}$ is equal to
A.) 3:2
B.) 1:3
C.) 1:2
D.) 2:3
Answer
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Hint: In this problem relation between the nuclear radius and atomic of an element according to which nucleus with radius R is related to mass number A that is given as $R = Ro{A^{\dfrac{1}{3}}}$ substitute the given values in the formula this step will help you to answer the given question.
Step By Step Answer:
As we know that there is a relation between the nuclear radius and atomic masses of an element. The relation says that the nuclear radius R is related to the mass number A as follows $R = Ro{A^{\dfrac{1}{3}}}$, here Ro=constant which value is $1.2 \times {10^{ - 15}}$ or $1.7 \times {10^{ - 15}}$
Thus, we can say that the nucleus of radius varies as ${A^{\dfrac{1}{3}}}$
While the atomic number increases so the electrons in the atom increases and the number of shells also increases.
Thus, size of the atom increases which means the radius of nucleus increases
According to this relation we can easily solve the question as we are given the mass number of radius 1 and radius 2 which is ${A_1}$ =216 and${A_2}$ = 64, and we have to find the ratio of radius 1 to that of radius 2
From the relation above ${R_1} = Ro{A^{\dfrac{1}{3}}}$ which will be ${R_1} = Ro \times {216^{\dfrac{1}{3}}}$
Similarly, ${R_2} = Ro{A^{\dfrac{1}{3}}}$ which will be ${R_2} = Ro \times {64^{\dfrac{1}{3}}}$
Now we can easily divide these two equations to find out their ratio:$$ $$
So, $\dfrac{{{R_1}}}{{{R_2}}}$= $\dfrac{{Ro \times {{216}^{\dfrac{1}{3}}}}}{{Ro \times {{64}^{\dfrac{1}{3}}}}}$
From here Ro and Ro gets cancel, so the remaining value is:
$ \Rightarrow $ $\dfrac{6}{4}$ =$\dfrac{3}{2}$ = 3: 2
Hence, option A is correct.
Note: A nucleus or nucleus contains neutrons, protons and electrons inside it where protons are positively charged collected particles and neutrons are electrically neutral whereas electrons are negatively charged particles. This nucleus or nucleus has a spherical shape because the nucleus projects strong nuclear force that binds these protons and neutrons in the nucleus together which just reaches a few proton diameter over a distance. Therefore, for the nuclei to maintain stability the nuclei tends to keep all of its dimension as small as possible.
Step By Step Answer:
As we know that there is a relation between the nuclear radius and atomic masses of an element. The relation says that the nuclear radius R is related to the mass number A as follows $R = Ro{A^{\dfrac{1}{3}}}$, here Ro=constant which value is $1.2 \times {10^{ - 15}}$ or $1.7 \times {10^{ - 15}}$
Thus, we can say that the nucleus of radius varies as ${A^{\dfrac{1}{3}}}$
While the atomic number increases so the electrons in the atom increases and the number of shells also increases.
Thus, size of the atom increases which means the radius of nucleus increases
According to this relation we can easily solve the question as we are given the mass number of radius 1 and radius 2 which is ${A_1}$ =216 and${A_2}$ = 64, and we have to find the ratio of radius 1 to that of radius 2
From the relation above ${R_1} = Ro{A^{\dfrac{1}{3}}}$ which will be ${R_1} = Ro \times {216^{\dfrac{1}{3}}}$
Similarly, ${R_2} = Ro{A^{\dfrac{1}{3}}}$ which will be ${R_2} = Ro \times {64^{\dfrac{1}{3}}}$
Now we can easily divide these two equations to find out their ratio:$$ $$
So, $\dfrac{{{R_1}}}{{{R_2}}}$= $\dfrac{{Ro \times {{216}^{\dfrac{1}{3}}}}}{{Ro \times {{64}^{\dfrac{1}{3}}}}}$
From here Ro and Ro gets cancel, so the remaining value is:
$ \Rightarrow $ $\dfrac{6}{4}$ =$\dfrac{3}{2}$ = 3: 2
Hence, option A is correct.
Note: A nucleus or nucleus contains neutrons, protons and electrons inside it where protons are positively charged collected particles and neutrons are electrically neutral whereas electrons are negatively charged particles. This nucleus or nucleus has a spherical shape because the nucleus projects strong nuclear force that binds these protons and neutrons in the nucleus together which just reaches a few proton diameter over a distance. Therefore, for the nuclei to maintain stability the nuclei tends to keep all of its dimension as small as possible.
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