147427
Atomic weight of boron is 10.81 and it has two isotopes ${ }_{5}^{10} B$ and ${ }_{5}^{11} B$. Then, the ratio of atoms of ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$ in nature would be
1 19:81
2 $10: 11$
3 $15: 16$
4 $81: 19$
Explanation:
A Given that, Atomic weight of boron $=10.81$ Let, the percentage of ${ }_{5} \mathrm{~B}^{10}$ atoms be $\mathrm{x}$ and percentage of ${ }_{5} \mathrm{~B}^{11}$ is $(100-\mathrm{x})$ then the average atomic weight, $\frac{10 x+11(100-x)}{100}=10.81$ $10 x+1100-11 x=1081$ $x=19$ So, percentage of ${ }_{5} \mathrm{~B}^{11}=100-19=81$ Thus, the required ratio is - $\frac{{ }_{5} \mathrm{~B}^{10}}{{ }_{5} \mathrm{~B}^{11}}=\frac{19}{81}$
AIPMT- 1998
NUCLEAR PHYSICS
147428
In compound $X(n, \alpha) \rightarrow{ }_{3} \mathrm{Li}^{7}$, the element $X$ is
1 ${ }_{2} \mathrm{He}^{4}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{4} \mathrm{~B}^{11}$
Explanation:
B ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{3} \mathrm{Li}^{7}+{ }_{2} \mathrm{He}^{4}$ For mass, $A+1=7+4$ $A=10$ For Atomic number $Z+0=3+2$ $Z=5$ Then, the element become ${ }_{5} \mathrm{X}^{10}={ }_{5} \mathrm{~B}^{10}$
AIPMT- 2001
NUCLEAR PHYSICS
147430
The volume occupied by an atom is greater than the volume of the nucleus by factor of about
1 $10^{10}$
2 $10^{15}$
3 $10^{1}$
4 $10^{5}$
Explanation:
B We know that, Radius of atom is of the order of $\left(r_{1}\right)=10^{-10} \mathrm{~m}$ Radius of the nucleus is of the order of $\left(\mathrm{r}_{2}\right)=10^{-15} \mathrm{~m}$ Then the ratio of required volume $=\frac{\text { Volume of atom }}{\text { Volume of nucleus }}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}{ }^{3}}$ $=\left(\frac{10^{-10}}{10^{-15}}\right)^{3}$ $=10^{15}$
AIPMT- 2003
NUCLEAR PHYSICS
147431
If radius of the ${ }_{13}^{27} \mathrm{Al}$ nucleus is taken to be $\mathbf{R}_{\mathrm{Al}}$, then the radius of ${ }_{53}^{125} \mathrm{Te}$ nucleus is nearly
B Given, atomic mass of $\mathrm{Al}$ and $\mathrm{Te}$ are $\mathrm{A}_{\mathrm{Al}}=27, \mathrm{~A}_{\mathrm{Te}}=125$, We know that, $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Where, $\mathrm{A}=$ mass number $\mathrm{R}=\text { radius of nucleus }$ Ratio of radius of $\mathrm{Te}$ and $\mathrm{Al}$ is $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\left(\frac{\mathrm{A}_{\mathrm{Te}}}{\mathrm{A}_{\mathrm{Al}}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{(125)^{1 / 3}}{(27)^{1 / 3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{5}{3}$ $\mathrm{R}_{\mathrm{Te}}=\frac{5}{3} \mathrm{R}_{\mathrm{Al}}$
147427
Atomic weight of boron is 10.81 and it has two isotopes ${ }_{5}^{10} B$ and ${ }_{5}^{11} B$. Then, the ratio of atoms of ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$ in nature would be
1 19:81
2 $10: 11$
3 $15: 16$
4 $81: 19$
Explanation:
A Given that, Atomic weight of boron $=10.81$ Let, the percentage of ${ }_{5} \mathrm{~B}^{10}$ atoms be $\mathrm{x}$ and percentage of ${ }_{5} \mathrm{~B}^{11}$ is $(100-\mathrm{x})$ then the average atomic weight, $\frac{10 x+11(100-x)}{100}=10.81$ $10 x+1100-11 x=1081$ $x=19$ So, percentage of ${ }_{5} \mathrm{~B}^{11}=100-19=81$ Thus, the required ratio is - $\frac{{ }_{5} \mathrm{~B}^{10}}{{ }_{5} \mathrm{~B}^{11}}=\frac{19}{81}$
AIPMT- 1998
NUCLEAR PHYSICS
147428
In compound $X(n, \alpha) \rightarrow{ }_{3} \mathrm{Li}^{7}$, the element $X$ is
1 ${ }_{2} \mathrm{He}^{4}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{4} \mathrm{~B}^{11}$
Explanation:
B ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{3} \mathrm{Li}^{7}+{ }_{2} \mathrm{He}^{4}$ For mass, $A+1=7+4$ $A=10$ For Atomic number $Z+0=3+2$ $Z=5$ Then, the element become ${ }_{5} \mathrm{X}^{10}={ }_{5} \mathrm{~B}^{10}$
AIPMT- 2001
NUCLEAR PHYSICS
147430
The volume occupied by an atom is greater than the volume of the nucleus by factor of about
1 $10^{10}$
2 $10^{15}$
3 $10^{1}$
4 $10^{5}$
Explanation:
B We know that, Radius of atom is of the order of $\left(r_{1}\right)=10^{-10} \mathrm{~m}$ Radius of the nucleus is of the order of $\left(\mathrm{r}_{2}\right)=10^{-15} \mathrm{~m}$ Then the ratio of required volume $=\frac{\text { Volume of atom }}{\text { Volume of nucleus }}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}{ }^{3}}$ $=\left(\frac{10^{-10}}{10^{-15}}\right)^{3}$ $=10^{15}$
AIPMT- 2003
NUCLEAR PHYSICS
147431
If radius of the ${ }_{13}^{27} \mathrm{Al}$ nucleus is taken to be $\mathbf{R}_{\mathrm{Al}}$, then the radius of ${ }_{53}^{125} \mathrm{Te}$ nucleus is nearly
B Given, atomic mass of $\mathrm{Al}$ and $\mathrm{Te}$ are $\mathrm{A}_{\mathrm{Al}}=27, \mathrm{~A}_{\mathrm{Te}}=125$, We know that, $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Where, $\mathrm{A}=$ mass number $\mathrm{R}=\text { radius of nucleus }$ Ratio of radius of $\mathrm{Te}$ and $\mathrm{Al}$ is $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\left(\frac{\mathrm{A}_{\mathrm{Te}}}{\mathrm{A}_{\mathrm{Al}}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{(125)^{1 / 3}}{(27)^{1 / 3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{5}{3}$ $\mathrm{R}_{\mathrm{Te}}=\frac{5}{3} \mathrm{R}_{\mathrm{Al}}$
147427
Atomic weight of boron is 10.81 and it has two isotopes ${ }_{5}^{10} B$ and ${ }_{5}^{11} B$. Then, the ratio of atoms of ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$ in nature would be
1 19:81
2 $10: 11$
3 $15: 16$
4 $81: 19$
Explanation:
A Given that, Atomic weight of boron $=10.81$ Let, the percentage of ${ }_{5} \mathrm{~B}^{10}$ atoms be $\mathrm{x}$ and percentage of ${ }_{5} \mathrm{~B}^{11}$ is $(100-\mathrm{x})$ then the average atomic weight, $\frac{10 x+11(100-x)}{100}=10.81$ $10 x+1100-11 x=1081$ $x=19$ So, percentage of ${ }_{5} \mathrm{~B}^{11}=100-19=81$ Thus, the required ratio is - $\frac{{ }_{5} \mathrm{~B}^{10}}{{ }_{5} \mathrm{~B}^{11}}=\frac{19}{81}$
AIPMT- 1998
NUCLEAR PHYSICS
147428
In compound $X(n, \alpha) \rightarrow{ }_{3} \mathrm{Li}^{7}$, the element $X$ is
1 ${ }_{2} \mathrm{He}^{4}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{4} \mathrm{~B}^{11}$
Explanation:
B ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{3} \mathrm{Li}^{7}+{ }_{2} \mathrm{He}^{4}$ For mass, $A+1=7+4$ $A=10$ For Atomic number $Z+0=3+2$ $Z=5$ Then, the element become ${ }_{5} \mathrm{X}^{10}={ }_{5} \mathrm{~B}^{10}$
AIPMT- 2001
NUCLEAR PHYSICS
147430
The volume occupied by an atom is greater than the volume of the nucleus by factor of about
1 $10^{10}$
2 $10^{15}$
3 $10^{1}$
4 $10^{5}$
Explanation:
B We know that, Radius of atom is of the order of $\left(r_{1}\right)=10^{-10} \mathrm{~m}$ Radius of the nucleus is of the order of $\left(\mathrm{r}_{2}\right)=10^{-15} \mathrm{~m}$ Then the ratio of required volume $=\frac{\text { Volume of atom }}{\text { Volume of nucleus }}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}{ }^{3}}$ $=\left(\frac{10^{-10}}{10^{-15}}\right)^{3}$ $=10^{15}$
AIPMT- 2003
NUCLEAR PHYSICS
147431
If radius of the ${ }_{13}^{27} \mathrm{Al}$ nucleus is taken to be $\mathbf{R}_{\mathrm{Al}}$, then the radius of ${ }_{53}^{125} \mathrm{Te}$ nucleus is nearly
B Given, atomic mass of $\mathrm{Al}$ and $\mathrm{Te}$ are $\mathrm{A}_{\mathrm{Al}}=27, \mathrm{~A}_{\mathrm{Te}}=125$, We know that, $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Where, $\mathrm{A}=$ mass number $\mathrm{R}=\text { radius of nucleus }$ Ratio of radius of $\mathrm{Te}$ and $\mathrm{Al}$ is $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\left(\frac{\mathrm{A}_{\mathrm{Te}}}{\mathrm{A}_{\mathrm{Al}}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{(125)^{1 / 3}}{(27)^{1 / 3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{5}{3}$ $\mathrm{R}_{\mathrm{Te}}=\frac{5}{3} \mathrm{R}_{\mathrm{Al}}$
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NUCLEAR PHYSICS
147427
Atomic weight of boron is 10.81 and it has two isotopes ${ }_{5}^{10} B$ and ${ }_{5}^{11} B$. Then, the ratio of atoms of ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$ in nature would be
1 19:81
2 $10: 11$
3 $15: 16$
4 $81: 19$
Explanation:
A Given that, Atomic weight of boron $=10.81$ Let, the percentage of ${ }_{5} \mathrm{~B}^{10}$ atoms be $\mathrm{x}$ and percentage of ${ }_{5} \mathrm{~B}^{11}$ is $(100-\mathrm{x})$ then the average atomic weight, $\frac{10 x+11(100-x)}{100}=10.81$ $10 x+1100-11 x=1081$ $x=19$ So, percentage of ${ }_{5} \mathrm{~B}^{11}=100-19=81$ Thus, the required ratio is - $\frac{{ }_{5} \mathrm{~B}^{10}}{{ }_{5} \mathrm{~B}^{11}}=\frac{19}{81}$
AIPMT- 1998
NUCLEAR PHYSICS
147428
In compound $X(n, \alpha) \rightarrow{ }_{3} \mathrm{Li}^{7}$, the element $X$ is
1 ${ }_{2} \mathrm{He}^{4}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{4} \mathrm{~B}^{11}$
Explanation:
B ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{3} \mathrm{Li}^{7}+{ }_{2} \mathrm{He}^{4}$ For mass, $A+1=7+4$ $A=10$ For Atomic number $Z+0=3+2$ $Z=5$ Then, the element become ${ }_{5} \mathrm{X}^{10}={ }_{5} \mathrm{~B}^{10}$
AIPMT- 2001
NUCLEAR PHYSICS
147430
The volume occupied by an atom is greater than the volume of the nucleus by factor of about
1 $10^{10}$
2 $10^{15}$
3 $10^{1}$
4 $10^{5}$
Explanation:
B We know that, Radius of atom is of the order of $\left(r_{1}\right)=10^{-10} \mathrm{~m}$ Radius of the nucleus is of the order of $\left(\mathrm{r}_{2}\right)=10^{-15} \mathrm{~m}$ Then the ratio of required volume $=\frac{\text { Volume of atom }}{\text { Volume of nucleus }}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}{ }^{3}}$ $=\left(\frac{10^{-10}}{10^{-15}}\right)^{3}$ $=10^{15}$
AIPMT- 2003
NUCLEAR PHYSICS
147431
If radius of the ${ }_{13}^{27} \mathrm{Al}$ nucleus is taken to be $\mathbf{R}_{\mathrm{Al}}$, then the radius of ${ }_{53}^{125} \mathrm{Te}$ nucleus is nearly
B Given, atomic mass of $\mathrm{Al}$ and $\mathrm{Te}$ are $\mathrm{A}_{\mathrm{Al}}=27, \mathrm{~A}_{\mathrm{Te}}=125$, We know that, $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Where, $\mathrm{A}=$ mass number $\mathrm{R}=\text { radius of nucleus }$ Ratio of radius of $\mathrm{Te}$ and $\mathrm{Al}$ is $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\left(\frac{\mathrm{A}_{\mathrm{Te}}}{\mathrm{A}_{\mathrm{Al}}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{(125)^{1 / 3}}{(27)^{1 / 3}}$ $\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\frac{5}{3}$ $\mathrm{R}_{\mathrm{Te}}=\frac{5}{3} \mathrm{R}_{\mathrm{Al}}$
