147362
When a uranium isotope ${ }_{92}^{235} \mathrm{U}$ is bombarded with a neutron, it generates ${ }_{36}^{89} \mathrm{Kr}$, three neutron and
1 ${ }_{40}^{91} \mathrm{Zr}$
2 ${ }_{36}^{101} \mathrm{Kr}$
3 ${ }_{36}^{103} \mathrm{Kr}$
4 ${ }_{56}^{144} \mathrm{Ba}$
Explanation:
D When a uranium isotope ${ }_{92} \mathrm{U}^{235}$ is bombarded with a neutron ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+3{ }_{0} \mathrm{n}^{1}+{ }_{36} \mathrm{Kr}^{89}+\mathrm{Q}($ Energy) Atomic mass in left hand side, $92+0=92$ Atomic number on the right hand side $36+3 \times 0+Q=36+Q$ Then the law of conservation of mass $36+\mathrm{Q}=92$ $\mathrm{Q}=56$ Atomic mass of left hand side $=$ Atomic mass of right hand side $235+1=89+3+\mathrm{P}$ $\mathrm{P}=144$ If, $\quad{ }_{56} \mathrm{X}^{144}$ then $\mathrm{X}=\mathrm{Ba}$ Compound will be ${ }_{56} \mathrm{Ba}^{144}$
NEET- (Sep) 2020
NUCLEAR PHYSICS
147364
Calculate the energy equivalent of $1 \mathrm{~g}$ of substance
1 $4 \times 10^{12} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $6 \times 10^{11} \mathrm{~J}$
4 $7 \times 10^{12} \mathrm{~J}$
Explanation:
B Given, $\operatorname{Mass}(\mathrm{m})=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ We know that, Energy $(\mathrm{E})=\mathrm{mc}^{2}$ $\mathrm{E}=\left(1 \times 10^{-3}\right)\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
GUJCET 2020
NUCLEAR PHYSICS
147366
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1$. What will be the ratio of their nuclear radius?
1 $2^{1 / 3}: 1$
2 $1: 2^{1 / 3}$
3 $3^{1 / 2}: 1$
4 $1: 3^{1 / 2}$
Explanation:
B Given that, velocity ratio are $\mathrm{v}_{1}: \mathrm{v}_{2}=2: 1$ From law of conservation of momentum, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho}{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{(2)^{1 / 3}}{1}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
BITSAT-2020
NUCLEAR PHYSICS
147367
$\alpha$-particle consists of
1 2 electrons, 2 protons and 2 neutrons
2 2 electrons and 4 protons only
3 2 protons only
4 2 protons and 2 neutrons only
Explanation:
D $\alpha$-particle is written as ${ }_{2} \mathrm{He}^{4}$. Atomic number $=$ Number of proton $=2$ Number of neutrons $=$ Mass number - Atomic number $=4-2$ $=2$ So, $\alpha$-particle is nucleus of Helium which has two protons and two neutrons.
NEET (national) - 2019
NUCLEAR PHYSICS
147369
Which of the following elements you need to remove to form an isotone family?
1 ${ }_{8} \mathrm{O}^{16}$
2 ${ }_{7} \mathrm{~N}^{15}$
3 ${ }_{6} \mathrm{C}^{14}$
4 ${ }_{13} \mathrm{Al}^{27}$
5 ${ }_{9} \mathrm{~F}^{17}$
Explanation:
D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ For ${ }_{8} \mathrm{O}^{16}=16-8=8$ For ${ }_{7} \mathrm{~N}^{15}=15-7=8$ For ${ }_{6} \mathrm{C}^{14}=14-6=8$ For ${ }_{9} \mathrm{~F}^{17}=17-9=8$ For ${ }_{13} \mathrm{~A}^{27}=27-13=14$ From this ${ }_{13} \mathrm{~A}^{27}$ has not same neutron number therefore, need to remove form an isotones family.
147362
When a uranium isotope ${ }_{92}^{235} \mathrm{U}$ is bombarded with a neutron, it generates ${ }_{36}^{89} \mathrm{Kr}$, three neutron and
1 ${ }_{40}^{91} \mathrm{Zr}$
2 ${ }_{36}^{101} \mathrm{Kr}$
3 ${ }_{36}^{103} \mathrm{Kr}$
4 ${ }_{56}^{144} \mathrm{Ba}$
Explanation:
D When a uranium isotope ${ }_{92} \mathrm{U}^{235}$ is bombarded with a neutron ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+3{ }_{0} \mathrm{n}^{1}+{ }_{36} \mathrm{Kr}^{89}+\mathrm{Q}($ Energy) Atomic mass in left hand side, $92+0=92$ Atomic number on the right hand side $36+3 \times 0+Q=36+Q$ Then the law of conservation of mass $36+\mathrm{Q}=92$ $\mathrm{Q}=56$ Atomic mass of left hand side $=$ Atomic mass of right hand side $235+1=89+3+\mathrm{P}$ $\mathrm{P}=144$ If, $\quad{ }_{56} \mathrm{X}^{144}$ then $\mathrm{X}=\mathrm{Ba}$ Compound will be ${ }_{56} \mathrm{Ba}^{144}$
NEET- (Sep) 2020
NUCLEAR PHYSICS
147364
Calculate the energy equivalent of $1 \mathrm{~g}$ of substance
1 $4 \times 10^{12} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $6 \times 10^{11} \mathrm{~J}$
4 $7 \times 10^{12} \mathrm{~J}$
Explanation:
B Given, $\operatorname{Mass}(\mathrm{m})=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ We know that, Energy $(\mathrm{E})=\mathrm{mc}^{2}$ $\mathrm{E}=\left(1 \times 10^{-3}\right)\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
GUJCET 2020
NUCLEAR PHYSICS
147366
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1$. What will be the ratio of their nuclear radius?
1 $2^{1 / 3}: 1$
2 $1: 2^{1 / 3}$
3 $3^{1 / 2}: 1$
4 $1: 3^{1 / 2}$
Explanation:
B Given that, velocity ratio are $\mathrm{v}_{1}: \mathrm{v}_{2}=2: 1$ From law of conservation of momentum, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho}{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{(2)^{1 / 3}}{1}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
BITSAT-2020
NUCLEAR PHYSICS
147367
$\alpha$-particle consists of
1 2 electrons, 2 protons and 2 neutrons
2 2 electrons and 4 protons only
3 2 protons only
4 2 protons and 2 neutrons only
Explanation:
D $\alpha$-particle is written as ${ }_{2} \mathrm{He}^{4}$. Atomic number $=$ Number of proton $=2$ Number of neutrons $=$ Mass number - Atomic number $=4-2$ $=2$ So, $\alpha$-particle is nucleus of Helium which has two protons and two neutrons.
NEET (national) - 2019
NUCLEAR PHYSICS
147369
Which of the following elements you need to remove to form an isotone family?
1 ${ }_{8} \mathrm{O}^{16}$
2 ${ }_{7} \mathrm{~N}^{15}$
3 ${ }_{6} \mathrm{C}^{14}$
4 ${ }_{13} \mathrm{Al}^{27}$
5 ${ }_{9} \mathrm{~F}^{17}$
Explanation:
D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ For ${ }_{8} \mathrm{O}^{16}=16-8=8$ For ${ }_{7} \mathrm{~N}^{15}=15-7=8$ For ${ }_{6} \mathrm{C}^{14}=14-6=8$ For ${ }_{9} \mathrm{~F}^{17}=17-9=8$ For ${ }_{13} \mathrm{~A}^{27}=27-13=14$ From this ${ }_{13} \mathrm{~A}^{27}$ has not same neutron number therefore, need to remove form an isotones family.
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NUCLEAR PHYSICS
147362
When a uranium isotope ${ }_{92}^{235} \mathrm{U}$ is bombarded with a neutron, it generates ${ }_{36}^{89} \mathrm{Kr}$, three neutron and
1 ${ }_{40}^{91} \mathrm{Zr}$
2 ${ }_{36}^{101} \mathrm{Kr}$
3 ${ }_{36}^{103} \mathrm{Kr}$
4 ${ }_{56}^{144} \mathrm{Ba}$
Explanation:
D When a uranium isotope ${ }_{92} \mathrm{U}^{235}$ is bombarded with a neutron ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+3{ }_{0} \mathrm{n}^{1}+{ }_{36} \mathrm{Kr}^{89}+\mathrm{Q}($ Energy) Atomic mass in left hand side, $92+0=92$ Atomic number on the right hand side $36+3 \times 0+Q=36+Q$ Then the law of conservation of mass $36+\mathrm{Q}=92$ $\mathrm{Q}=56$ Atomic mass of left hand side $=$ Atomic mass of right hand side $235+1=89+3+\mathrm{P}$ $\mathrm{P}=144$ If, $\quad{ }_{56} \mathrm{X}^{144}$ then $\mathrm{X}=\mathrm{Ba}$ Compound will be ${ }_{56} \mathrm{Ba}^{144}$
NEET- (Sep) 2020
NUCLEAR PHYSICS
147364
Calculate the energy equivalent of $1 \mathrm{~g}$ of substance
1 $4 \times 10^{12} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $6 \times 10^{11} \mathrm{~J}$
4 $7 \times 10^{12} \mathrm{~J}$
Explanation:
B Given, $\operatorname{Mass}(\mathrm{m})=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ We know that, Energy $(\mathrm{E})=\mathrm{mc}^{2}$ $\mathrm{E}=\left(1 \times 10^{-3}\right)\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
GUJCET 2020
NUCLEAR PHYSICS
147366
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1$. What will be the ratio of their nuclear radius?
1 $2^{1 / 3}: 1$
2 $1: 2^{1 / 3}$
3 $3^{1 / 2}: 1$
4 $1: 3^{1 / 2}$
Explanation:
B Given that, velocity ratio are $\mathrm{v}_{1}: \mathrm{v}_{2}=2: 1$ From law of conservation of momentum, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho}{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{(2)^{1 / 3}}{1}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
BITSAT-2020
NUCLEAR PHYSICS
147367
$\alpha$-particle consists of
1 2 electrons, 2 protons and 2 neutrons
2 2 electrons and 4 protons only
3 2 protons only
4 2 protons and 2 neutrons only
Explanation:
D $\alpha$-particle is written as ${ }_{2} \mathrm{He}^{4}$. Atomic number $=$ Number of proton $=2$ Number of neutrons $=$ Mass number - Atomic number $=4-2$ $=2$ So, $\alpha$-particle is nucleus of Helium which has two protons and two neutrons.
NEET (national) - 2019
NUCLEAR PHYSICS
147369
Which of the following elements you need to remove to form an isotone family?
1 ${ }_{8} \mathrm{O}^{16}$
2 ${ }_{7} \mathrm{~N}^{15}$
3 ${ }_{6} \mathrm{C}^{14}$
4 ${ }_{13} \mathrm{Al}^{27}$
5 ${ }_{9} \mathrm{~F}^{17}$
Explanation:
D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ For ${ }_{8} \mathrm{O}^{16}=16-8=8$ For ${ }_{7} \mathrm{~N}^{15}=15-7=8$ For ${ }_{6} \mathrm{C}^{14}=14-6=8$ For ${ }_{9} \mathrm{~F}^{17}=17-9=8$ For ${ }_{13} \mathrm{~A}^{27}=27-13=14$ From this ${ }_{13} \mathrm{~A}^{27}$ has not same neutron number therefore, need to remove form an isotones family.
147362
When a uranium isotope ${ }_{92}^{235} \mathrm{U}$ is bombarded with a neutron, it generates ${ }_{36}^{89} \mathrm{Kr}$, three neutron and
1 ${ }_{40}^{91} \mathrm{Zr}$
2 ${ }_{36}^{101} \mathrm{Kr}$
3 ${ }_{36}^{103} \mathrm{Kr}$
4 ${ }_{56}^{144} \mathrm{Ba}$
Explanation:
D When a uranium isotope ${ }_{92} \mathrm{U}^{235}$ is bombarded with a neutron ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+3{ }_{0} \mathrm{n}^{1}+{ }_{36} \mathrm{Kr}^{89}+\mathrm{Q}($ Energy) Atomic mass in left hand side, $92+0=92$ Atomic number on the right hand side $36+3 \times 0+Q=36+Q$ Then the law of conservation of mass $36+\mathrm{Q}=92$ $\mathrm{Q}=56$ Atomic mass of left hand side $=$ Atomic mass of right hand side $235+1=89+3+\mathrm{P}$ $\mathrm{P}=144$ If, $\quad{ }_{56} \mathrm{X}^{144}$ then $\mathrm{X}=\mathrm{Ba}$ Compound will be ${ }_{56} \mathrm{Ba}^{144}$
NEET- (Sep) 2020
NUCLEAR PHYSICS
147364
Calculate the energy equivalent of $1 \mathrm{~g}$ of substance
1 $4 \times 10^{12} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $6 \times 10^{11} \mathrm{~J}$
4 $7 \times 10^{12} \mathrm{~J}$
Explanation:
B Given, $\operatorname{Mass}(\mathrm{m})=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ We know that, Energy $(\mathrm{E})=\mathrm{mc}^{2}$ $\mathrm{E}=\left(1 \times 10^{-3}\right)\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
GUJCET 2020
NUCLEAR PHYSICS
147366
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1$. What will be the ratio of their nuclear radius?
1 $2^{1 / 3}: 1$
2 $1: 2^{1 / 3}$
3 $3^{1 / 2}: 1$
4 $1: 3^{1 / 2}$
Explanation:
B Given that, velocity ratio are $\mathrm{v}_{1}: \mathrm{v}_{2}=2: 1$ From law of conservation of momentum, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho}{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{(2)^{1 / 3}}{1}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
BITSAT-2020
NUCLEAR PHYSICS
147367
$\alpha$-particle consists of
1 2 electrons, 2 protons and 2 neutrons
2 2 electrons and 4 protons only
3 2 protons only
4 2 protons and 2 neutrons only
Explanation:
D $\alpha$-particle is written as ${ }_{2} \mathrm{He}^{4}$. Atomic number $=$ Number of proton $=2$ Number of neutrons $=$ Mass number - Atomic number $=4-2$ $=2$ So, $\alpha$-particle is nucleus of Helium which has two protons and two neutrons.
NEET (national) - 2019
NUCLEAR PHYSICS
147369
Which of the following elements you need to remove to form an isotone family?
1 ${ }_{8} \mathrm{O}^{16}$
2 ${ }_{7} \mathrm{~N}^{15}$
3 ${ }_{6} \mathrm{C}^{14}$
4 ${ }_{13} \mathrm{Al}^{27}$
5 ${ }_{9} \mathrm{~F}^{17}$
Explanation:
D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ For ${ }_{8} \mathrm{O}^{16}=16-8=8$ For ${ }_{7} \mathrm{~N}^{15}=15-7=8$ For ${ }_{6} \mathrm{C}^{14}=14-6=8$ For ${ }_{9} \mathrm{~F}^{17}=17-9=8$ For ${ }_{13} \mathrm{~A}^{27}=27-13=14$ From this ${ }_{13} \mathrm{~A}^{27}$ has not same neutron number therefore, need to remove form an isotones family.
147362
When a uranium isotope ${ }_{92}^{235} \mathrm{U}$ is bombarded with a neutron, it generates ${ }_{36}^{89} \mathrm{Kr}$, three neutron and
1 ${ }_{40}^{91} \mathrm{Zr}$
2 ${ }_{36}^{101} \mathrm{Kr}$
3 ${ }_{36}^{103} \mathrm{Kr}$
4 ${ }_{56}^{144} \mathrm{Ba}$
Explanation:
D When a uranium isotope ${ }_{92} \mathrm{U}^{235}$ is bombarded with a neutron ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+3{ }_{0} \mathrm{n}^{1}+{ }_{36} \mathrm{Kr}^{89}+\mathrm{Q}($ Energy) Atomic mass in left hand side, $92+0=92$ Atomic number on the right hand side $36+3 \times 0+Q=36+Q$ Then the law of conservation of mass $36+\mathrm{Q}=92$ $\mathrm{Q}=56$ Atomic mass of left hand side $=$ Atomic mass of right hand side $235+1=89+3+\mathrm{P}$ $\mathrm{P}=144$ If, $\quad{ }_{56} \mathrm{X}^{144}$ then $\mathrm{X}=\mathrm{Ba}$ Compound will be ${ }_{56} \mathrm{Ba}^{144}$
NEET- (Sep) 2020
NUCLEAR PHYSICS
147364
Calculate the energy equivalent of $1 \mathrm{~g}$ of substance
1 $4 \times 10^{12} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $6 \times 10^{11} \mathrm{~J}$
4 $7 \times 10^{12} \mathrm{~J}$
Explanation:
B Given, $\operatorname{Mass}(\mathrm{m})=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ We know that, Energy $(\mathrm{E})=\mathrm{mc}^{2}$ $\mathrm{E}=\left(1 \times 10^{-3}\right)\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
GUJCET 2020
NUCLEAR PHYSICS
147366
A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1$. What will be the ratio of their nuclear radius?
1 $2^{1 / 3}: 1$
2 $1: 2^{1 / 3}$
3 $3^{1 / 2}: 1$
4 $1: 3^{1 / 2}$
Explanation:
B Given that, velocity ratio are $\mathrm{v}_{1}: \mathrm{v}_{2}=2: 1$ From law of conservation of momentum, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho}{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3}$ $\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{(2)^{1 / 3}}{1}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
BITSAT-2020
NUCLEAR PHYSICS
147367
$\alpha$-particle consists of
1 2 electrons, 2 protons and 2 neutrons
2 2 electrons and 4 protons only
3 2 protons only
4 2 protons and 2 neutrons only
Explanation:
D $\alpha$-particle is written as ${ }_{2} \mathrm{He}^{4}$. Atomic number $=$ Number of proton $=2$ Number of neutrons $=$ Mass number - Atomic number $=4-2$ $=2$ So, $\alpha$-particle is nucleus of Helium which has two protons and two neutrons.
NEET (national) - 2019
NUCLEAR PHYSICS
147369
Which of the following elements you need to remove to form an isotone family?
1 ${ }_{8} \mathrm{O}^{16}$
2 ${ }_{7} \mathrm{~N}^{15}$
3 ${ }_{6} \mathrm{C}^{14}$
4 ${ }_{13} \mathrm{Al}^{27}$
5 ${ }_{9} \mathrm{~F}^{17}$
Explanation:
D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ For ${ }_{8} \mathrm{O}^{16}=16-8=8$ For ${ }_{7} \mathrm{~N}^{15}=15-7=8$ For ${ }_{6} \mathrm{C}^{14}=14-6=8$ For ${ }_{9} \mathrm{~F}^{17}=17-9=8$ For ${ }_{13} \mathrm{~A}^{27}=27-13=14$ From this ${ }_{13} \mathrm{~A}^{27}$ has not same neutron number therefore, need to remove form an isotones family.