147403
If the radius of a nucleus of mass number 3 is $R$, then the radius of a nucleus of mass number 81 is
1 $3 \mathrm{R}$
2 $9 \mathrm{R}$
3 $(27)^{1 / 2} \mathrm{R}$
4 $27 \mathrm{R}$
Explanation:
A Given, atomic mass are $\mathrm{A}_{1}=3, \mathrm{~A}_{2}=81$ and radius of nucleus are $R_{1}=R, R_{2}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Ratio of radius of nucleus, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{3}{81}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}=\left(\frac{1}{3}\right)^{3 \times 1 / 3}$ $\mathrm{R}_{2}=3 \mathrm{R}$
J and K CET- 2010
NUCLEAR PHYSICS
147407
The ratio of radii of nuclei of two atoms of elements of atomic mass numbers 27 and 64 is
1 $3: 4$
2 $4: 3$
3 $9: 16$
4 $16: 9$
Explanation:
A Given, Atomic mass numbers are $A_{1}=27$ and $A_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Ratio of radii of nuclei $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ Then, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{1 / 3}=\left(\frac{3}{4}\right)^{3 \times 1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}$ So, $\quad \mathrm{R}_{1}: \mathrm{R}_{2}=3: 4$
J and K CET-2014
NUCLEAR PHYSICS
147408
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is
1 $5: 4$
2 $25: 16$
3 $1: 1$
4 $4: 5$
Explanation:
A Given that, $\mathrm{A}_{1}=125, \mathrm{~A}_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{125}{64}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{5}{4}\right)$ $\mathrm{R}_{1}: \mathrm{R}_{2}=5: 4$
NUCLEAR PHYSICS
147411
Which of the following particles do not exist in the ${ }_{92} \mathrm{U}^{238}$ nucleus?
1 92 protons
2 92 electrons
3 146 neutrons
4 none of these
Explanation:
B Atomic representation $={ }_{92} \mathrm{U}^{238}$ Number of proton $=92$ mass number $=238$ mass number $=$ number of proton - number of neutrons number of neutron $=238-92$ $=146$ Electron do not exist in the nucleus thus, nucleus comprise of proton and neutron.
147403
If the radius of a nucleus of mass number 3 is $R$, then the radius of a nucleus of mass number 81 is
1 $3 \mathrm{R}$
2 $9 \mathrm{R}$
3 $(27)^{1 / 2} \mathrm{R}$
4 $27 \mathrm{R}$
Explanation:
A Given, atomic mass are $\mathrm{A}_{1}=3, \mathrm{~A}_{2}=81$ and radius of nucleus are $R_{1}=R, R_{2}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Ratio of radius of nucleus, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{3}{81}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}=\left(\frac{1}{3}\right)^{3 \times 1 / 3}$ $\mathrm{R}_{2}=3 \mathrm{R}$
J and K CET- 2010
NUCLEAR PHYSICS
147407
The ratio of radii of nuclei of two atoms of elements of atomic mass numbers 27 and 64 is
1 $3: 4$
2 $4: 3$
3 $9: 16$
4 $16: 9$
Explanation:
A Given, Atomic mass numbers are $A_{1}=27$ and $A_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Ratio of radii of nuclei $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ Then, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{1 / 3}=\left(\frac{3}{4}\right)^{3 \times 1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}$ So, $\quad \mathrm{R}_{1}: \mathrm{R}_{2}=3: 4$
J and K CET-2014
NUCLEAR PHYSICS
147408
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is
1 $5: 4$
2 $25: 16$
3 $1: 1$
4 $4: 5$
Explanation:
A Given that, $\mathrm{A}_{1}=125, \mathrm{~A}_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{125}{64}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{5}{4}\right)$ $\mathrm{R}_{1}: \mathrm{R}_{2}=5: 4$
NUCLEAR PHYSICS
147411
Which of the following particles do not exist in the ${ }_{92} \mathrm{U}^{238}$ nucleus?
1 92 protons
2 92 electrons
3 146 neutrons
4 none of these
Explanation:
B Atomic representation $={ }_{92} \mathrm{U}^{238}$ Number of proton $=92$ mass number $=238$ mass number $=$ number of proton - number of neutrons number of neutron $=238-92$ $=146$ Electron do not exist in the nucleus thus, nucleus comprise of proton and neutron.
147403
If the radius of a nucleus of mass number 3 is $R$, then the radius of a nucleus of mass number 81 is
1 $3 \mathrm{R}$
2 $9 \mathrm{R}$
3 $(27)^{1 / 2} \mathrm{R}$
4 $27 \mathrm{R}$
Explanation:
A Given, atomic mass are $\mathrm{A}_{1}=3, \mathrm{~A}_{2}=81$ and radius of nucleus are $R_{1}=R, R_{2}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Ratio of radius of nucleus, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{3}{81}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}=\left(\frac{1}{3}\right)^{3 \times 1 / 3}$ $\mathrm{R}_{2}=3 \mathrm{R}$
J and K CET- 2010
NUCLEAR PHYSICS
147407
The ratio of radii of nuclei of two atoms of elements of atomic mass numbers 27 and 64 is
1 $3: 4$
2 $4: 3$
3 $9: 16$
4 $16: 9$
Explanation:
A Given, Atomic mass numbers are $A_{1}=27$ and $A_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Ratio of radii of nuclei $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ Then, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{1 / 3}=\left(\frac{3}{4}\right)^{3 \times 1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}$ So, $\quad \mathrm{R}_{1}: \mathrm{R}_{2}=3: 4$
J and K CET-2014
NUCLEAR PHYSICS
147408
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is
1 $5: 4$
2 $25: 16$
3 $1: 1$
4 $4: 5$
Explanation:
A Given that, $\mathrm{A}_{1}=125, \mathrm{~A}_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{125}{64}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{5}{4}\right)$ $\mathrm{R}_{1}: \mathrm{R}_{2}=5: 4$
NUCLEAR PHYSICS
147411
Which of the following particles do not exist in the ${ }_{92} \mathrm{U}^{238}$ nucleus?
1 92 protons
2 92 electrons
3 146 neutrons
4 none of these
Explanation:
B Atomic representation $={ }_{92} \mathrm{U}^{238}$ Number of proton $=92$ mass number $=238$ mass number $=$ number of proton - number of neutrons number of neutron $=238-92$ $=146$ Electron do not exist in the nucleus thus, nucleus comprise of proton and neutron.
147403
If the radius of a nucleus of mass number 3 is $R$, then the radius of a nucleus of mass number 81 is
1 $3 \mathrm{R}$
2 $9 \mathrm{R}$
3 $(27)^{1 / 2} \mathrm{R}$
4 $27 \mathrm{R}$
Explanation:
A Given, atomic mass are $\mathrm{A}_{1}=3, \mathrm{~A}_{2}=81$ and radius of nucleus are $R_{1}=R, R_{2}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Ratio of radius of nucleus, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{3}{81}\right)^{1 / 3}$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}=\left(\frac{1}{3}\right)^{3 \times 1 / 3}$ $\mathrm{R}_{2}=3 \mathrm{R}$
J and K CET- 2010
NUCLEAR PHYSICS
147407
The ratio of radii of nuclei of two atoms of elements of atomic mass numbers 27 and 64 is
1 $3: 4$
2 $4: 3$
3 $9: 16$
4 $16: 9$
Explanation:
A Given, Atomic mass numbers are $A_{1}=27$ and $A_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Ratio of radii of nuclei $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ Then, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{1 / 3}=\left(\frac{3}{4}\right)^{3 \times 1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}$ So, $\quad \mathrm{R}_{1}: \mathrm{R}_{2}=3: 4$
J and K CET-2014
NUCLEAR PHYSICS
147408
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is
1 $5: 4$
2 $25: 16$
3 $1: 1$
4 $4: 5$
Explanation:
A Given that, $\mathrm{A}_{1}=125, \mathrm{~A}_{2}=64$ We know that, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{125}{64}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{5}{4}\right)$ $\mathrm{R}_{1}: \mathrm{R}_{2}=5: 4$
NUCLEAR PHYSICS
147411
Which of the following particles do not exist in the ${ }_{92} \mathrm{U}^{238}$ nucleus?
1 92 protons
2 92 electrons
3 146 neutrons
4 none of these
Explanation:
B Atomic representation $={ }_{92} \mathrm{U}^{238}$ Number of proton $=92$ mass number $=238$ mass number $=$ number of proton - number of neutrons number of neutron $=238-92$ $=146$ Electron do not exist in the nucleus thus, nucleus comprise of proton and neutron.
