145268
If $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei of mass numbers 64 and 125 respectively, then the ratio $\left(r_{1} / r_{2}\right)$ is
1 $\frac{64}{125}$
2 $\sqrt{\frac{64}{125}}$
3 $\frac{5}{4}$
4 $\frac{4}{5}$
Explanation:
D Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=125, \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=$ ? Where $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei. We know that, $r=r_{0}(A)^{1 / 3} \quad \text { (Where, } r_{0}=\text { constant) }$ $r \propto A^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{64}{125}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\frac{4}{5}$
JCECE-2009
ATOMS
145274
If the ratio of radii of nuclei ${ }_{13}^{27} \mathrm{Al}$ and ${ }_{20}^{\mathrm{A}} \mathrm{X}$ is 3 $: 5$, ten the number of neutrons in the nuclei of $\mathrm{X}$ will be-
1 13
2 52
3 100
4 73
Explanation:
D Given, mass of ${ }_{13} \mathrm{~A} l^{27}\left(\mathrm{~A}_{1}\right)=27$, mass of ${ }_{52} \mathrm{X}^{\mathrm{A}}\left(\mathrm{A}_{2}\right)=2, \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{\mathrm{~A}}\right)^{1 / 3} \Rightarrow \frac{3}{5}=\frac{3}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=5 \Rightarrow 5^{3}=125$ Number of neutron in the nuclei of $\mathrm{X}=\mathrm{A}-52 \Rightarrow 125-$ $52=73$
BCECE-2012
ATOMS
145275
Given mass number of gold $=197$, Density of gold $=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $=6 \times 10^{23}$. The radius of the gold atom is approximately
1 $1.5 \times 10^{-8} \mathrm{~m}$
2 $1.7 \times 10^{-9} \mathrm{~m}$
3 $1.5 \times 10^{-10} \mathrm{~m}$
4 $1.5 \times 10^{-12} \mathrm{~m}$
Explanation:
C Given, Mass number of gold $(\mathrm{m})=197$ a.m.u Density of gold $(\rho)=1.97 \mathrm{~g} / \mathrm{cm}^{3}=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23}$ Volume occupied by one gram atom of gold $=\frac{m}{\rho}$ $=\frac{197 \mathrm{~g}}{19.7 \mathrm{~g} / \mathrm{cm}^{3}}=10 \mathrm{~cm}^{3}$ Volume of one atom $=\frac{10}{\mathrm{~N}_{\mathrm{A}}}=\frac{10}{56 \times 10^{23}}=\frac{5}{3} \times 10^{-23} \mathrm{~cm}^{3}$ Let $\mathrm{r}$ be the radius of the atom $\therefore \quad \frac{4}{3} \pi \mathrm{r}^{3}=\frac{5 \times 10^{-23}}{3}$ $\mathrm{r}^{3}=\frac{5 \times 10^{-23}}{4 \times 3.14}$ $\mathrm{r} =1.5 \times 10^{-10} \mathrm{~m}$
BCECE-2011
ATOMS
145277
An electron jumps from the first excited state to the ground state of hydrogen atom. What will be the percentage change in the speed of electron?
1 $25 \%$
2 $50 \%$
3 $100 \%$
4 $200 \%$
Explanation:
B Velocity of electron in any state $\mathrm{n}$ $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{kZ \textrm {e } ^ { 2 }}}{\mathrm{nh}}$ For $\mathrm{n}=1$ come to ground state, $\mathrm{Z}=1$ (Hydrogen atom) $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{\mathrm{~h}}$ For $\mathrm{n}=2$ $\mathrm{V}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{2 \mathrm{~h}}$ On dividing equation (ii) by equation (i) will get as follows $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $1-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=1-\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}}=\frac{2-1}{2}=\frac{1}{2}$ $\Delta \mathrm{v} \%=\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}} \times 100=\frac{1}{2} \times 100=50 \%$
145268
If $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei of mass numbers 64 and 125 respectively, then the ratio $\left(r_{1} / r_{2}\right)$ is
1 $\frac{64}{125}$
2 $\sqrt{\frac{64}{125}}$
3 $\frac{5}{4}$
4 $\frac{4}{5}$
Explanation:
D Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=125, \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=$ ? Where $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei. We know that, $r=r_{0}(A)^{1 / 3} \quad \text { (Where, } r_{0}=\text { constant) }$ $r \propto A^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{64}{125}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\frac{4}{5}$
JCECE-2009
ATOMS
145274
If the ratio of radii of nuclei ${ }_{13}^{27} \mathrm{Al}$ and ${ }_{20}^{\mathrm{A}} \mathrm{X}$ is 3 $: 5$, ten the number of neutrons in the nuclei of $\mathrm{X}$ will be-
1 13
2 52
3 100
4 73
Explanation:
D Given, mass of ${ }_{13} \mathrm{~A} l^{27}\left(\mathrm{~A}_{1}\right)=27$, mass of ${ }_{52} \mathrm{X}^{\mathrm{A}}\left(\mathrm{A}_{2}\right)=2, \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{\mathrm{~A}}\right)^{1 / 3} \Rightarrow \frac{3}{5}=\frac{3}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=5 \Rightarrow 5^{3}=125$ Number of neutron in the nuclei of $\mathrm{X}=\mathrm{A}-52 \Rightarrow 125-$ $52=73$
BCECE-2012
ATOMS
145275
Given mass number of gold $=197$, Density of gold $=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $=6 \times 10^{23}$. The radius of the gold atom is approximately
1 $1.5 \times 10^{-8} \mathrm{~m}$
2 $1.7 \times 10^{-9} \mathrm{~m}$
3 $1.5 \times 10^{-10} \mathrm{~m}$
4 $1.5 \times 10^{-12} \mathrm{~m}$
Explanation:
C Given, Mass number of gold $(\mathrm{m})=197$ a.m.u Density of gold $(\rho)=1.97 \mathrm{~g} / \mathrm{cm}^{3}=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23}$ Volume occupied by one gram atom of gold $=\frac{m}{\rho}$ $=\frac{197 \mathrm{~g}}{19.7 \mathrm{~g} / \mathrm{cm}^{3}}=10 \mathrm{~cm}^{3}$ Volume of one atom $=\frac{10}{\mathrm{~N}_{\mathrm{A}}}=\frac{10}{56 \times 10^{23}}=\frac{5}{3} \times 10^{-23} \mathrm{~cm}^{3}$ Let $\mathrm{r}$ be the radius of the atom $\therefore \quad \frac{4}{3} \pi \mathrm{r}^{3}=\frac{5 \times 10^{-23}}{3}$ $\mathrm{r}^{3}=\frac{5 \times 10^{-23}}{4 \times 3.14}$ $\mathrm{r} =1.5 \times 10^{-10} \mathrm{~m}$
BCECE-2011
ATOMS
145277
An electron jumps from the first excited state to the ground state of hydrogen atom. What will be the percentage change in the speed of electron?
1 $25 \%$
2 $50 \%$
3 $100 \%$
4 $200 \%$
Explanation:
B Velocity of electron in any state $\mathrm{n}$ $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{kZ \textrm {e } ^ { 2 }}}{\mathrm{nh}}$ For $\mathrm{n}=1$ come to ground state, $\mathrm{Z}=1$ (Hydrogen atom) $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{\mathrm{~h}}$ For $\mathrm{n}=2$ $\mathrm{V}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{2 \mathrm{~h}}$ On dividing equation (ii) by equation (i) will get as follows $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $1-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=1-\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}}=\frac{2-1}{2}=\frac{1}{2}$ $\Delta \mathrm{v} \%=\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}} \times 100=\frac{1}{2} \times 100=50 \%$
145268
If $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei of mass numbers 64 and 125 respectively, then the ratio $\left(r_{1} / r_{2}\right)$ is
1 $\frac{64}{125}$
2 $\sqrt{\frac{64}{125}}$
3 $\frac{5}{4}$
4 $\frac{4}{5}$
Explanation:
D Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=125, \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=$ ? Where $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei. We know that, $r=r_{0}(A)^{1 / 3} \quad \text { (Where, } r_{0}=\text { constant) }$ $r \propto A^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{64}{125}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\frac{4}{5}$
JCECE-2009
ATOMS
145274
If the ratio of radii of nuclei ${ }_{13}^{27} \mathrm{Al}$ and ${ }_{20}^{\mathrm{A}} \mathrm{X}$ is 3 $: 5$, ten the number of neutrons in the nuclei of $\mathrm{X}$ will be-
1 13
2 52
3 100
4 73
Explanation:
D Given, mass of ${ }_{13} \mathrm{~A} l^{27}\left(\mathrm{~A}_{1}\right)=27$, mass of ${ }_{52} \mathrm{X}^{\mathrm{A}}\left(\mathrm{A}_{2}\right)=2, \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{\mathrm{~A}}\right)^{1 / 3} \Rightarrow \frac{3}{5}=\frac{3}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=5 \Rightarrow 5^{3}=125$ Number of neutron in the nuclei of $\mathrm{X}=\mathrm{A}-52 \Rightarrow 125-$ $52=73$
BCECE-2012
ATOMS
145275
Given mass number of gold $=197$, Density of gold $=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $=6 \times 10^{23}$. The radius of the gold atom is approximately
1 $1.5 \times 10^{-8} \mathrm{~m}$
2 $1.7 \times 10^{-9} \mathrm{~m}$
3 $1.5 \times 10^{-10} \mathrm{~m}$
4 $1.5 \times 10^{-12} \mathrm{~m}$
Explanation:
C Given, Mass number of gold $(\mathrm{m})=197$ a.m.u Density of gold $(\rho)=1.97 \mathrm{~g} / \mathrm{cm}^{3}=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23}$ Volume occupied by one gram atom of gold $=\frac{m}{\rho}$ $=\frac{197 \mathrm{~g}}{19.7 \mathrm{~g} / \mathrm{cm}^{3}}=10 \mathrm{~cm}^{3}$ Volume of one atom $=\frac{10}{\mathrm{~N}_{\mathrm{A}}}=\frac{10}{56 \times 10^{23}}=\frac{5}{3} \times 10^{-23} \mathrm{~cm}^{3}$ Let $\mathrm{r}$ be the radius of the atom $\therefore \quad \frac{4}{3} \pi \mathrm{r}^{3}=\frac{5 \times 10^{-23}}{3}$ $\mathrm{r}^{3}=\frac{5 \times 10^{-23}}{4 \times 3.14}$ $\mathrm{r} =1.5 \times 10^{-10} \mathrm{~m}$
BCECE-2011
ATOMS
145277
An electron jumps from the first excited state to the ground state of hydrogen atom. What will be the percentage change in the speed of electron?
1 $25 \%$
2 $50 \%$
3 $100 \%$
4 $200 \%$
Explanation:
B Velocity of electron in any state $\mathrm{n}$ $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{kZ \textrm {e } ^ { 2 }}}{\mathrm{nh}}$ For $\mathrm{n}=1$ come to ground state, $\mathrm{Z}=1$ (Hydrogen atom) $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{\mathrm{~h}}$ For $\mathrm{n}=2$ $\mathrm{V}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{2 \mathrm{~h}}$ On dividing equation (ii) by equation (i) will get as follows $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $1-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=1-\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}}=\frac{2-1}{2}=\frac{1}{2}$ $\Delta \mathrm{v} \%=\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}} \times 100=\frac{1}{2} \times 100=50 \%$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145268
If $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei of mass numbers 64 and 125 respectively, then the ratio $\left(r_{1} / r_{2}\right)$ is
1 $\frac{64}{125}$
2 $\sqrt{\frac{64}{125}}$
3 $\frac{5}{4}$
4 $\frac{4}{5}$
Explanation:
D Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=125, \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=$ ? Where $r_{1}$ and $r_{2}$ are the radii of the atomic nuclei. We know that, $r=r_{0}(A)^{1 / 3} \quad \text { (Where, } r_{0}=\text { constant) }$ $r \propto A^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\left(\frac{64}{125}\right)^{1 / 3}$ $\frac{r_{1}}{r_{2}}=\frac{4}{5}$
JCECE-2009
ATOMS
145274
If the ratio of radii of nuclei ${ }_{13}^{27} \mathrm{Al}$ and ${ }_{20}^{\mathrm{A}} \mathrm{X}$ is 3 $: 5$, ten the number of neutrons in the nuclei of $\mathrm{X}$ will be-
1 13
2 52
3 100
4 73
Explanation:
D Given, mass of ${ }_{13} \mathrm{~A} l^{27}\left(\mathrm{~A}_{1}\right)=27$, mass of ${ }_{52} \mathrm{X}^{\mathrm{A}}\left(\mathrm{A}_{2}\right)=2, \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{\mathrm{~A}}\right)^{1 / 3} \Rightarrow \frac{3}{5}=\frac{3}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=5 \Rightarrow 5^{3}=125$ Number of neutron in the nuclei of $\mathrm{X}=\mathrm{A}-52 \Rightarrow 125-$ $52=73$
BCECE-2012
ATOMS
145275
Given mass number of gold $=197$, Density of gold $=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $=6 \times 10^{23}$. The radius of the gold atom is approximately
1 $1.5 \times 10^{-8} \mathrm{~m}$
2 $1.7 \times 10^{-9} \mathrm{~m}$
3 $1.5 \times 10^{-10} \mathrm{~m}$
4 $1.5 \times 10^{-12} \mathrm{~m}$
Explanation:
C Given, Mass number of gold $(\mathrm{m})=197$ a.m.u Density of gold $(\rho)=1.97 \mathrm{~g} / \mathrm{cm}^{3}=19.7 \mathrm{~g} / \mathrm{cm}^{3}$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23}$ Volume occupied by one gram atom of gold $=\frac{m}{\rho}$ $=\frac{197 \mathrm{~g}}{19.7 \mathrm{~g} / \mathrm{cm}^{3}}=10 \mathrm{~cm}^{3}$ Volume of one atom $=\frac{10}{\mathrm{~N}_{\mathrm{A}}}=\frac{10}{56 \times 10^{23}}=\frac{5}{3} \times 10^{-23} \mathrm{~cm}^{3}$ Let $\mathrm{r}$ be the radius of the atom $\therefore \quad \frac{4}{3} \pi \mathrm{r}^{3}=\frac{5 \times 10^{-23}}{3}$ $\mathrm{r}^{3}=\frac{5 \times 10^{-23}}{4 \times 3.14}$ $\mathrm{r} =1.5 \times 10^{-10} \mathrm{~m}$
BCECE-2011
ATOMS
145277
An electron jumps from the first excited state to the ground state of hydrogen atom. What will be the percentage change in the speed of electron?
1 $25 \%$
2 $50 \%$
3 $100 \%$
4 $200 \%$
Explanation:
B Velocity of electron in any state $\mathrm{n}$ $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{kZ \textrm {e } ^ { 2 }}}{\mathrm{nh}}$ For $\mathrm{n}=1$ come to ground state, $\mathrm{Z}=1$ (Hydrogen atom) $\mathrm{v}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{\mathrm{~h}}$ For $\mathrm{n}=2$ $\mathrm{V}_{\mathrm{n}}=\frac{2 \pi \mathrm{ke}^{2}}{2 \mathrm{~h}}$ On dividing equation (ii) by equation (i) will get as follows $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $1-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=1-\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{n}_{2}}$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}}=\frac{2-1}{2}=\frac{1}{2}$ $\Delta \mathrm{v} \%=\frac{\Delta \mathrm{v}}{\mathrm{v}_{1}} \times 100=\frac{1}{2} \times 100=50 \%$
