147352
Mass numbers of two nuclei are in the ratio of 4: 3. Their nuclear densities will be in the ratio of
1 $4: 3$
2 $\left(\frac{3}{4}\right)^{1 / 3}$
3 $1: 1$
4 $\left(\frac{4}{3}\right)^{1 / 3}$
Explanation:
C Given that, $\mathrm{A}_{1}: \mathrm{A}_{2}=4: 3$ We know that- $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Density of nucleus $=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho=\frac{m \times A}{\frac{4}{3} \pi R_{0}^{3} \times A}$ $\rho=\frac{3 m}{4 \pi R_{0}{ }^{3}}$ The density of nucleus is independent of mass number(A). Therefore the density of nucleus is the same for every element and hence, the ratio will be $1: 1$
JEE Main-26.07.2022
NUCLEAR PHYSICS
147353
Which of the following figure represents the variation of $\left(\frac{R}{R_{0}}\right)$ with in $A$ (if $R=$ radius of a nucleus and $A=$ its mass number) (a)
1
2
3
4
Explanation:
B We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{\frac{1}{3}}$ Taking $\log$ on both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ It represent a straight line.
JEE Main-25.06.2022
NUCLEAR PHYSICS
147354
What is the mass number of the nucleus having radius equal to $\frac{1}{3}$ of that of ${ }^{189} \mathrm{Os}$ ?
1 20
2 7
3 12
4 14 TS
Explanation:
B Given, Radius of nucleus $(R)=\frac{1}{3} R_{O s}$, Atomic mass of Os (A) $=189$ We know that, Bohr radius Then, $\quad \mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{\mathrm{Os}}=\mathrm{R}_{0}(189)^{1 / 3}$ equation (ii) divided by equation (i) we get- $\frac{\mathrm{R}_{\mathrm{OS}}}{\mathrm{R}_{\mathrm{Os}} / 3}=[189 / \mathrm{A}]^{1 / 3}$ $(3)^{3}=\frac{189}{\mathrm{~A}}$ $\therefore \quad \mathrm{A}=\frac{189}{27}$ Mass number of the nucleus (A) $=7$
EAMCET 19.07.2022
NUCLEAR PHYSICS
147355
Estimate the approximate volume of aluminum nucleus $(A=27)$ Use $(\mathbf{R}_{\mathbf{0}} = \mathbf{1 . 0 \times 1 0 ^ { - 1 5 }} \mathrm{m}$ $\boldsymbol{\pi} = \mathbf{3}$
147352
Mass numbers of two nuclei are in the ratio of 4: 3. Their nuclear densities will be in the ratio of
1 $4: 3$
2 $\left(\frac{3}{4}\right)^{1 / 3}$
3 $1: 1$
4 $\left(\frac{4}{3}\right)^{1 / 3}$
Explanation:
C Given that, $\mathrm{A}_{1}: \mathrm{A}_{2}=4: 3$ We know that- $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Density of nucleus $=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho=\frac{m \times A}{\frac{4}{3} \pi R_{0}^{3} \times A}$ $\rho=\frac{3 m}{4 \pi R_{0}{ }^{3}}$ The density of nucleus is independent of mass number(A). Therefore the density of nucleus is the same for every element and hence, the ratio will be $1: 1$
JEE Main-26.07.2022
NUCLEAR PHYSICS
147353
Which of the following figure represents the variation of $\left(\frac{R}{R_{0}}\right)$ with in $A$ (if $R=$ radius of a nucleus and $A=$ its mass number) (a)
1
2
3
4
Explanation:
B We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{\frac{1}{3}}$ Taking $\log$ on both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ It represent a straight line.
JEE Main-25.06.2022
NUCLEAR PHYSICS
147354
What is the mass number of the nucleus having radius equal to $\frac{1}{3}$ of that of ${ }^{189} \mathrm{Os}$ ?
1 20
2 7
3 12
4 14 TS
Explanation:
B Given, Radius of nucleus $(R)=\frac{1}{3} R_{O s}$, Atomic mass of Os (A) $=189$ We know that, Bohr radius Then, $\quad \mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{\mathrm{Os}}=\mathrm{R}_{0}(189)^{1 / 3}$ equation (ii) divided by equation (i) we get- $\frac{\mathrm{R}_{\mathrm{OS}}}{\mathrm{R}_{\mathrm{Os}} / 3}=[189 / \mathrm{A}]^{1 / 3}$ $(3)^{3}=\frac{189}{\mathrm{~A}}$ $\therefore \quad \mathrm{A}=\frac{189}{27}$ Mass number of the nucleus (A) $=7$
EAMCET 19.07.2022
NUCLEAR PHYSICS
147355
Estimate the approximate volume of aluminum nucleus $(A=27)$ Use $(\mathbf{R}_{\mathbf{0}} = \mathbf{1 . 0 \times 1 0 ^ { - 1 5 }} \mathrm{m}$ $\boldsymbol{\pi} = \mathbf{3}$
147352
Mass numbers of two nuclei are in the ratio of 4: 3. Their nuclear densities will be in the ratio of
1 $4: 3$
2 $\left(\frac{3}{4}\right)^{1 / 3}$
3 $1: 1$
4 $\left(\frac{4}{3}\right)^{1 / 3}$
Explanation:
C Given that, $\mathrm{A}_{1}: \mathrm{A}_{2}=4: 3$ We know that- $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Density of nucleus $=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho=\frac{m \times A}{\frac{4}{3} \pi R_{0}^{3} \times A}$ $\rho=\frac{3 m}{4 \pi R_{0}{ }^{3}}$ The density of nucleus is independent of mass number(A). Therefore the density of nucleus is the same for every element and hence, the ratio will be $1: 1$
JEE Main-26.07.2022
NUCLEAR PHYSICS
147353
Which of the following figure represents the variation of $\left(\frac{R}{R_{0}}\right)$ with in $A$ (if $R=$ radius of a nucleus and $A=$ its mass number) (a)
1
2
3
4
Explanation:
B We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{\frac{1}{3}}$ Taking $\log$ on both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ It represent a straight line.
JEE Main-25.06.2022
NUCLEAR PHYSICS
147354
What is the mass number of the nucleus having radius equal to $\frac{1}{3}$ of that of ${ }^{189} \mathrm{Os}$ ?
1 20
2 7
3 12
4 14 TS
Explanation:
B Given, Radius of nucleus $(R)=\frac{1}{3} R_{O s}$, Atomic mass of Os (A) $=189$ We know that, Bohr radius Then, $\quad \mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{\mathrm{Os}}=\mathrm{R}_{0}(189)^{1 / 3}$ equation (ii) divided by equation (i) we get- $\frac{\mathrm{R}_{\mathrm{OS}}}{\mathrm{R}_{\mathrm{Os}} / 3}=[189 / \mathrm{A}]^{1 / 3}$ $(3)^{3}=\frac{189}{\mathrm{~A}}$ $\therefore \quad \mathrm{A}=\frac{189}{27}$ Mass number of the nucleus (A) $=7$
EAMCET 19.07.2022
NUCLEAR PHYSICS
147355
Estimate the approximate volume of aluminum nucleus $(A=27)$ Use $(\mathbf{R}_{\mathbf{0}} = \mathbf{1 . 0 \times 1 0 ^ { - 1 5 }} \mathrm{m}$ $\boldsymbol{\pi} = \mathbf{3}$
147352
Mass numbers of two nuclei are in the ratio of 4: 3. Their nuclear densities will be in the ratio of
1 $4: 3$
2 $\left(\frac{3}{4}\right)^{1 / 3}$
3 $1: 1$
4 $\left(\frac{4}{3}\right)^{1 / 3}$
Explanation:
C Given that, $\mathrm{A}_{1}: \mathrm{A}_{2}=4: 3$ We know that- $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Density of nucleus $=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho=\frac{m \times A}{\frac{4}{3} \pi R_{0}^{3} \times A}$ $\rho=\frac{3 m}{4 \pi R_{0}{ }^{3}}$ The density of nucleus is independent of mass number(A). Therefore the density of nucleus is the same for every element and hence, the ratio will be $1: 1$
JEE Main-26.07.2022
NUCLEAR PHYSICS
147353
Which of the following figure represents the variation of $\left(\frac{R}{R_{0}}\right)$ with in $A$ (if $R=$ radius of a nucleus and $A=$ its mass number) (a)
1
2
3
4
Explanation:
B We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{\frac{1}{3}}$ Taking $\log$ on both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ It represent a straight line.
JEE Main-25.06.2022
NUCLEAR PHYSICS
147354
What is the mass number of the nucleus having radius equal to $\frac{1}{3}$ of that of ${ }^{189} \mathrm{Os}$ ?
1 20
2 7
3 12
4 14 TS
Explanation:
B Given, Radius of nucleus $(R)=\frac{1}{3} R_{O s}$, Atomic mass of Os (A) $=189$ We know that, Bohr radius Then, $\quad \mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{\mathrm{Os}}=\mathrm{R}_{0}(189)^{1 / 3}$ equation (ii) divided by equation (i) we get- $\frac{\mathrm{R}_{\mathrm{OS}}}{\mathrm{R}_{\mathrm{Os}} / 3}=[189 / \mathrm{A}]^{1 / 3}$ $(3)^{3}=\frac{189}{\mathrm{~A}}$ $\therefore \quad \mathrm{A}=\frac{189}{27}$ Mass number of the nucleus (A) $=7$
EAMCET 19.07.2022
NUCLEAR PHYSICS
147355
Estimate the approximate volume of aluminum nucleus $(A=27)$ Use $(\mathbf{R}_{\mathbf{0}} = \mathbf{1 . 0 \times 1 0 ^ { - 1 5 }} \mathrm{m}$ $\boldsymbol{\pi} = \mathbf{3}$
