QBManager

NUCLEAR PHYSICS

147362 When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} Kr$ , three neutron and

1

_{40}^{91} Zr

2

_{36}^{101} Kr

3

_{36}^{103} Kr

4

_{56}^{144} Ba

Explanation:

D When a uranium isotope $_{92} U^{235}$ is bombarded with a neutron
$_{92} U^{235} +_{0} n^{1} \to_{Q} X^{P} + 3_{0} n^{1} +_{36} {Kr}^{89} + Q ($ 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
$36 + Q = 92$
$Q = 56$ Atomic mass of left hand side $=$ Atomic mass of right hand side
$235 + 1 = 89 + 3 + P$
$P = 144$
If, $_{56} X^{144}$ then $X = Ba$
Compound will be $_{56} {Ba}^{144}$

NEET- (Sep) 2020

NUCLEAR PHYSICS

147364 Calculate the energy equivalent of $1 g$ of substance

1

4 \times 10^{12} J

2

9 \times 10^{13} J

3

6 \times 10^{11} J

4

7 \times 10^{12} J

Explanation:

B Given,
$Mass (m) = 1 g = 1 \times 10^{- 3} kg$
Speed $(c) = 3 \times 10^{8} m / \sec$
We know that,
Energy $(E) = {mc}^{2}$
$E = (1 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$E = 9 \times 10^{13} J$

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 $v_{1} : v_{2} = 2 : 1$
From law of conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{\frac{4}{3} π r_{2}^{3} ρ}{\frac{4}{3} π r_{1}^{3} ρ} = {(\frac{r_{2}}{r_{1}})}^{3}$
$\frac{r_{2}}{r_{1}} = \frac{(2)^{1 / 3}}{1}$
$r_{1} : r_{2} = 1 : 2^{1 / 3}$

BITSAT-2020

NUCLEAR PHYSICS

147367 $α$ -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 $α$ -particle is written as $_{2} {He}^{4}$ .
Atomic number $=$ Number of proton $= 2$
Number of neutrons $=$ Mass number - Atomic number
$= 4 - 2$
$= 2$
So, $α$ -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} O^{16}

2

_{7} N^{15}

3

_{6} C^{14}

4

_{13} {Al}^{27}

5

_{9} F^{17}

Explanation:

D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(Z)$
For $_{8} O^{16} = 16 - 8 = 8$
For $_{7} N^{15} = 15 - 7 = 8$
For $_{6} C^{14} = 14 - 6 = 8$
For $_{9} F^{17} = 17 - 9 = 8$
For $_{13} A^{27} = 27 - 13 = 14$
From this $_{13} A^{27}$ has not same neutron number therefore, need to remove form an isotones family.

Kerala CEE-2019

NUCLEAR PHYSICS

147362 When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} Kr$ , three neutron and

1

_{40}^{91} Zr

2

_{36}^{101} Kr

3

_{36}^{103} Kr

4

_{56}^{144} Ba

Explanation:

D When a uranium isotope $_{92} U^{235}$ is bombarded with a neutron
$_{92} U^{235} +_{0} n^{1} \to_{Q} X^{P} + 3_{0} n^{1} +_{36} {Kr}^{89} + Q ($ 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
$36 + Q = 92$
$Q = 56$ Atomic mass of left hand side $=$ Atomic mass of right hand side
$235 + 1 = 89 + 3 + P$
$P = 144$
If, $_{56} X^{144}$ then $X = Ba$
Compound will be $_{56} {Ba}^{144}$

NEET- (Sep) 2020

NUCLEAR PHYSICS

147364 Calculate the energy equivalent of $1 g$ of substance

1

4 \times 10^{12} J

2

9 \times 10^{13} J

3

6 \times 10^{11} J

4

7 \times 10^{12} J

Explanation:

B Given,
$Mass (m) = 1 g = 1 \times 10^{- 3} kg$
Speed $(c) = 3 \times 10^{8} m / \sec$
We know that,
Energy $(E) = {mc}^{2}$
$E = (1 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$E = 9 \times 10^{13} J$

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 $v_{1} : v_{2} = 2 : 1$
From law of conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{\frac{4}{3} π r_{2}^{3} ρ}{\frac{4}{3} π r_{1}^{3} ρ} = {(\frac{r_{2}}{r_{1}})}^{3}$
$\frac{r_{2}}{r_{1}} = \frac{(2)^{1 / 3}}{1}$
$r_{1} : r_{2} = 1 : 2^{1 / 3}$

BITSAT-2020

NUCLEAR PHYSICS

147367 $α$ -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 $α$ -particle is written as $_{2} {He}^{4}$ .
Atomic number $=$ Number of proton $= 2$
Number of neutrons $=$ Mass number - Atomic number
$= 4 - 2$
$= 2$
So, $α$ -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} O^{16}

2

_{7} N^{15}

3

_{6} C^{14}

4

_{13} {Al}^{27}

5

_{9} F^{17}

Explanation:

D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(Z)$
For $_{8} O^{16} = 16 - 8 = 8$
For $_{7} N^{15} = 15 - 7 = 8$
For $_{6} C^{14} = 14 - 6 = 8$
For $_{9} F^{17} = 17 - 9 = 8$
For $_{13} A^{27} = 27 - 13 = 14$
From this $_{13} A^{27}$ has not same neutron number therefore, need to remove form an isotones family.

Kerala CEE-2019

NUCLEAR PHYSICS

147362 When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} Kr$ , three neutron and

1

_{40}^{91} Zr

2

_{36}^{101} Kr

3

_{36}^{103} Kr

4

_{56}^{144} Ba

Explanation:

D When a uranium isotope $_{92} U^{235}$ is bombarded with a neutron
$_{92} U^{235} +_{0} n^{1} \to_{Q} X^{P} + 3_{0} n^{1} +_{36} {Kr}^{89} + Q ($ 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
$36 + Q = 92$
$Q = 56$ Atomic mass of left hand side $=$ Atomic mass of right hand side
$235 + 1 = 89 + 3 + P$
$P = 144$
If, $_{56} X^{144}$ then $X = Ba$
Compound will be $_{56} {Ba}^{144}$

NEET- (Sep) 2020

NUCLEAR PHYSICS

147364 Calculate the energy equivalent of $1 g$ of substance

1

4 \times 10^{12} J

2

9 \times 10^{13} J

3

6 \times 10^{11} J

4

7 \times 10^{12} J

Explanation:

B Given,
$Mass (m) = 1 g = 1 \times 10^{- 3} kg$
Speed $(c) = 3 \times 10^{8} m / \sec$
We know that,
Energy $(E) = {mc}^{2}$
$E = (1 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$E = 9 \times 10^{13} J$

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 $v_{1} : v_{2} = 2 : 1$
From law of conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{\frac{4}{3} π r_{2}^{3} ρ}{\frac{4}{3} π r_{1}^{3} ρ} = {(\frac{r_{2}}{r_{1}})}^{3}$
$\frac{r_{2}}{r_{1}} = \frac{(2)^{1 / 3}}{1}$
$r_{1} : r_{2} = 1 : 2^{1 / 3}$

BITSAT-2020

NUCLEAR PHYSICS

147367 $α$ -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 $α$ -particle is written as $_{2} {He}^{4}$ .
Atomic number $=$ Number of proton $= 2$
Number of neutrons $=$ Mass number - Atomic number
$= 4 - 2$
$= 2$
So, $α$ -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} O^{16}

2

_{7} N^{15}

3

_{6} C^{14}

4

_{13} {Al}^{27}

5

_{9} F^{17}

Explanation:

D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(Z)$
For $_{8} O^{16} = 16 - 8 = 8$
For $_{7} N^{15} = 15 - 7 = 8$
For $_{6} C^{14} = 14 - 6 = 8$
For $_{9} F^{17} = 17 - 9 = 8$
For $_{13} A^{27} = 27 - 13 = 14$
From this $_{13} A^{27}$ has not same neutron number therefore, need to remove form an isotones family.

Kerala CEE-2019

NUCLEAR PHYSICS

147362 When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} Kr$ , three neutron and

1

_{40}^{91} Zr

2

_{36}^{101} Kr

3

_{36}^{103} Kr

4

_{56}^{144} Ba

Explanation:

D When a uranium isotope $_{92} U^{235}$ is bombarded with a neutron
$_{92} U^{235} +_{0} n^{1} \to_{Q} X^{P} + 3_{0} n^{1} +_{36} {Kr}^{89} + Q ($ 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
$36 + Q = 92$
$Q = 56$ Atomic mass of left hand side $=$ Atomic mass of right hand side
$235 + 1 = 89 + 3 + P$
$P = 144$
If, $_{56} X^{144}$ then $X = Ba$
Compound will be $_{56} {Ba}^{144}$

NEET- (Sep) 2020

NUCLEAR PHYSICS

147364 Calculate the energy equivalent of $1 g$ of substance

1

4 \times 10^{12} J

2

9 \times 10^{13} J

3

6 \times 10^{11} J

4

7 \times 10^{12} J

Explanation:

B Given,
$Mass (m) = 1 g = 1 \times 10^{- 3} kg$
Speed $(c) = 3 \times 10^{8} m / \sec$
We know that,
Energy $(E) = {mc}^{2}$
$E = (1 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$E = 9 \times 10^{13} J$

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 $v_{1} : v_{2} = 2 : 1$
From law of conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{\frac{4}{3} π r_{2}^{3} ρ}{\frac{4}{3} π r_{1}^{3} ρ} = {(\frac{r_{2}}{r_{1}})}^{3}$
$\frac{r_{2}}{r_{1}} = \frac{(2)^{1 / 3}}{1}$
$r_{1} : r_{2} = 1 : 2^{1 / 3}$

BITSAT-2020

NUCLEAR PHYSICS

147367 $α$ -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 $α$ -particle is written as $_{2} {He}^{4}$ .
Atomic number $=$ Number of proton $= 2$
Number of neutrons $=$ Mass number - Atomic number
$= 4 - 2$
$= 2$
So, $α$ -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} O^{16}

2

_{7} N^{15}

3

_{6} C^{14}

4

_{13} {Al}^{27}

5

_{9} F^{17}

Explanation:

D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(Z)$
For $_{8} O^{16} = 16 - 8 = 8$
For $_{7} N^{15} = 15 - 7 = 8$
For $_{6} C^{14} = 14 - 6 = 8$
For $_{9} F^{17} = 17 - 9 = 8$
For $_{13} A^{27} = 27 - 13 = 14$
From this $_{13} A^{27}$ has not same neutron number therefore, need to remove form an isotones family.

Kerala CEE-2019

NUCLEAR PHYSICS

147362 When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} Kr$ , three neutron and

1

_{40}^{91} Zr

2

_{36}^{101} Kr

3

_{36}^{103} Kr

4

_{56}^{144} Ba

Explanation:

D When a uranium isotope $_{92} U^{235}$ is bombarded with a neutron
$_{92} U^{235} +_{0} n^{1} \to_{Q} X^{P} + 3_{0} n^{1} +_{36} {Kr}^{89} + Q ($ 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
$36 + Q = 92$
$Q = 56$ Atomic mass of left hand side $=$ Atomic mass of right hand side
$235 + 1 = 89 + 3 + P$
$P = 144$
If, $_{56} X^{144}$ then $X = Ba$
Compound will be $_{56} {Ba}^{144}$

NEET- (Sep) 2020

NUCLEAR PHYSICS

147364 Calculate the energy equivalent of $1 g$ of substance

1

4 \times 10^{12} J

2

9 \times 10^{13} J

3

6 \times 10^{11} J

4

7 \times 10^{12} J

Explanation:

B Given,
$Mass (m) = 1 g = 1 \times 10^{- 3} kg$
Speed $(c) = 3 \times 10^{8} m / \sec$
We know that,
Energy $(E) = {mc}^{2}$
$E = (1 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$E = 9 \times 10^{13} J$

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 $v_{1} : v_{2} = 2 : 1$
From law of conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{\frac{4}{3} π r_{2}^{3} ρ}{\frac{4}{3} π r_{1}^{3} ρ} = {(\frac{r_{2}}{r_{1}})}^{3}$
$\frac{r_{2}}{r_{1}} = \frac{(2)^{1 / 3}}{1}$
$r_{1} : r_{2} = 1 : 2^{1 / 3}$

BITSAT-2020

NUCLEAR PHYSICS

147367 $α$ -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 $α$ -particle is written as $_{2} {He}^{4}$ .
Atomic number $=$ Number of proton $= 2$
Number of neutrons $=$ Mass number - Atomic number
$= 4 - 2$
$= 2$
So, $α$ -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} O^{16}

2

_{7} N^{15}

3

_{6} C^{14}

4

_{13} {Al}^{27}

5

_{9} F^{17}

Explanation:

D Isotones means number of neutron is same Number of neutron $=$ Mass number (A) - Atomic number $(Z)$
For $_{8} O^{16} = 16 - 8 = 8$
For $_{7} N^{15} = 15 - 7 = 8$
For $_{6} C^{14} = 14 - 6 = 8$
For $_{9} F^{17} = 17 - 9 = 8$
For $_{13} A^{27} = 27 - 13 = 14$
From this $_{13} A^{27}$ has not same neutron number therefore, need to remove form an isotones family.

Kerala CEE-2019

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