QBManager

NUCLEAR PHYSICS

147394 The number of neutrons in $_{92} U^{235}$ nucleus is

1 327

2 235

3 143

4 92

Explanation:

C Given that,
Mass number $(A) = 235$
Number of protons $(Z) = 92$
Then number of neutrons $(N) = A - Z$
$N = 235 - 92$
$N = 143$

COMEDK 2012

NUCLEAR PHYSICS

147396 In $_{88} {R a}^{226}$ nucleus there are :

1 226 protons and 88 electrons

2 138 protons and 88 neutrons

3 226 neutrons and 138 electrons

4 138 neutrons and 88 protons

Explanation:

D Given that, $_{88} {Ra}^{226}$
Atomic number $(Z) = 88$
Mass number $(A) = 226$
Number of neutrons $= A - Z$
$= 226 - 88$
$= 138$
Number of proton is same as atomic number i.e. 88

AIIMS-1998

NUCLEAR PHYSICS

147397 What is the ratio of the radii of the nuclei of $_{13} {Al}^{27}$ and $_{52} {Te}^{125}$ ?

1

3 : 5

2

1 : 4

3

27 : 125

4

5 : 3

Explanation:

A We know that,
$R = R_{0} A^{1 / 3}$
Then, ratio of radius of $Al$ and $Te -$
$\frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} \Rightarrow \frac{R_{Al}}{R_{Te}} = {(\frac{3^{3}}{5^{3}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = \frac{3}{5}$
$R_{Al} : R_{Te} = 3 : 5$

SCRA-2011

NUCLEAR PHYSICS

147395 Find the energy equivalent of one atomic mass unit in joules and in $M e V$ .
#[Qdiff: Hard, QCat: Numerical Based, examname: AIIMS-2001]
, $1 AMU = 1.6605 \times 10^{- 27} kg, c = 3 \times 10^{8} m / s$
, $E = {mc}^{2}$
, $E = 1.6605 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$
, $E = 1.5 \times 10^{- 10} J$
, Energy in $MeV$
, $E = \frac{1.5 \times 10^{- 10} J}{1.6 \times 10^{- 19}} eV$
, $E = 0.9315 \times 10^{9} eV$
, $E = 931.5 \times 10^{6} eV$
, $E = 931.5 MeV$
, 57. Which one of the following has the highest neutrons ratio?
, (a) $_{92} U^{235}$
, (c) $_{2} {He}^{4}$
, (b) $_{8} O^{16}$
, (d) $_{26} {Fe}^{56}$
]#

1

1.66 \times 10^{- 10} J, 93.15 M e V

2

2.5 \times 10^{- 10} J, 931.5 M e V

3

1.5 \times 10^{- 10} J, 931.5 Me V

4

3 \times 10^{- 10} J, 931.5 M e V

Explanation:

C Ans: a : Neutrons ratio $= \frac{number of neutrons}{number of protons} = \frac{n}{p}$
For $_{92} U^{235}, n = 235 - 92 = 143$
Neutrons ratio $= \frac{143}{92} = 1.55$
For $_{8} O^{16} -$
$n = 16 - 8 = 8$
$p = 8$
Neutrons ratio $= \frac{8}{8} = 1$
For $_{2} {He}^{4} -$
$n = 4 - 2 = 2$
$p = 2$
Neutrons ratio $= \frac{2}{2} = 1$
For $_{26} {Fe}^{56} -$
$n = 56 - 26 = 30$
$p = 26$
Neutrons ratio $= \frac{30}{26} = 1.15$
So, highest neutrons ratio 1.55 for $_{92} U^{235}$ .

NUCLEAR PHYSICS

147402 If elements with principal quantum number $n$ $> 4$ were not allowed in nature, the number of possible elements would have been :

1 32

2 60

3 64

4 4

Explanation:

B We know,
Number of elements $= 2 n^{2}$
Then,
For, $n = 1$
Number of elements $= 2$
For, $n = 2$
Number of elements $= 8$
For, $n = 3$
Number of elements $= 18$
For, $n = 4$
Number of elements $= 32$
So, total number of element $= 32 + 18 + 8 + 2 = 60$

Karnataka CET-2003

NUCLEAR PHYSICS

147394 The number of neutrons in $_{92} U^{235}$ nucleus is

1 327

2 235

3 143

4 92

Explanation:

C Given that,
Mass number $(A) = 235$
Number of protons $(Z) = 92$
Then number of neutrons $(N) = A - Z$
$N = 235 - 92$
$N = 143$

COMEDK 2012

NUCLEAR PHYSICS

147396 In $_{88} {R a}^{226}$ nucleus there are :

1 226 protons and 88 electrons

2 138 protons and 88 neutrons

3 226 neutrons and 138 electrons

4 138 neutrons and 88 protons

Explanation:

D Given that, $_{88} {Ra}^{226}$
Atomic number $(Z) = 88$
Mass number $(A) = 226$
Number of neutrons $= A - Z$
$= 226 - 88$
$= 138$
Number of proton is same as atomic number i.e. 88

AIIMS-1998

NUCLEAR PHYSICS

147397 What is the ratio of the radii of the nuclei of $_{13} {Al}^{27}$ and $_{52} {Te}^{125}$ ?

1

3 : 5

2

1 : 4

3

27 : 125

4

5 : 3

Explanation:

A We know that,
$R = R_{0} A^{1 / 3}$
Then, ratio of radius of $Al$ and $Te -$
$\frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} \Rightarrow \frac{R_{Al}}{R_{Te}} = {(\frac{3^{3}}{5^{3}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = \frac{3}{5}$
$R_{Al} : R_{Te} = 3 : 5$

SCRA-2011

NUCLEAR PHYSICS

147395 Find the energy equivalent of one atomic mass unit in joules and in $M e V$ .
#[Qdiff: Hard, QCat: Numerical Based, examname: AIIMS-2001]
, $1 AMU = 1.6605 \times 10^{- 27} kg, c = 3 \times 10^{8} m / s$
, $E = {mc}^{2}$
, $E = 1.6605 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$
, $E = 1.5 \times 10^{- 10} J$
, Energy in $MeV$
, $E = \frac{1.5 \times 10^{- 10} J}{1.6 \times 10^{- 19}} eV$
, $E = 0.9315 \times 10^{9} eV$
, $E = 931.5 \times 10^{6} eV$
, $E = 931.5 MeV$
, 57. Which one of the following has the highest neutrons ratio?
, (a) $_{92} U^{235}$
, (c) $_{2} {He}^{4}$
, (b) $_{8} O^{16}$
, (d) $_{26} {Fe}^{56}$
]#

1

1.66 \times 10^{- 10} J, 93.15 M e V

2

2.5 \times 10^{- 10} J, 931.5 M e V

3

1.5 \times 10^{- 10} J, 931.5 Me V

4

3 \times 10^{- 10} J, 931.5 M e V

Explanation:

C Ans: a : Neutrons ratio $= \frac{number of neutrons}{number of protons} = \frac{n}{p}$
For $_{92} U^{235}, n = 235 - 92 = 143$
Neutrons ratio $= \frac{143}{92} = 1.55$
For $_{8} O^{16} -$
$n = 16 - 8 = 8$
$p = 8$
Neutrons ratio $= \frac{8}{8} = 1$
For $_{2} {He}^{4} -$
$n = 4 - 2 = 2$
$p = 2$
Neutrons ratio $= \frac{2}{2} = 1$
For $_{26} {Fe}^{56} -$
$n = 56 - 26 = 30$
$p = 26$
Neutrons ratio $= \frac{30}{26} = 1.15$
So, highest neutrons ratio 1.55 for $_{92} U^{235}$ .

NUCLEAR PHYSICS

147402 If elements with principal quantum number $n$ $> 4$ were not allowed in nature, the number of possible elements would have been :

1 32

2 60

3 64

4 4

Explanation:

B We know,
Number of elements $= 2 n^{2}$
Then,
For, $n = 1$
Number of elements $= 2$
For, $n = 2$
Number of elements $= 8$
For, $n = 3$
Number of elements $= 18$
For, $n = 4$
Number of elements $= 32$
So, total number of element $= 32 + 18 + 8 + 2 = 60$

Karnataka CET-2003

NUCLEAR PHYSICS

147394 The number of neutrons in $_{92} U^{235}$ nucleus is

1 327

2 235

3 143

4 92

Explanation:

C Given that,
Mass number $(A) = 235$
Number of protons $(Z) = 92$
Then number of neutrons $(N) = A - Z$
$N = 235 - 92$
$N = 143$

COMEDK 2012

NUCLEAR PHYSICS

147396 In $_{88} {R a}^{226}$ nucleus there are :

1 226 protons and 88 electrons

2 138 protons and 88 neutrons

3 226 neutrons and 138 electrons

4 138 neutrons and 88 protons

Explanation:

D Given that, $_{88} {Ra}^{226}$
Atomic number $(Z) = 88$
Mass number $(A) = 226$
Number of neutrons $= A - Z$
$= 226 - 88$
$= 138$
Number of proton is same as atomic number i.e. 88

AIIMS-1998

NUCLEAR PHYSICS

147397 What is the ratio of the radii of the nuclei of $_{13} {Al}^{27}$ and $_{52} {Te}^{125}$ ?

1

3 : 5

2

1 : 4

3

27 : 125

4

5 : 3

Explanation:

A We know that,
$R = R_{0} A^{1 / 3}$
Then, ratio of radius of $Al$ and $Te -$
$\frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} \Rightarrow \frac{R_{Al}}{R_{Te}} = {(\frac{3^{3}}{5^{3}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = \frac{3}{5}$
$R_{Al} : R_{Te} = 3 : 5$

SCRA-2011

NUCLEAR PHYSICS

147395 Find the energy equivalent of one atomic mass unit in joules and in $M e V$ .
#[Qdiff: Hard, QCat: Numerical Based, examname: AIIMS-2001]
, $1 AMU = 1.6605 \times 10^{- 27} kg, c = 3 \times 10^{8} m / s$
, $E = {mc}^{2}$
, $E = 1.6605 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$
, $E = 1.5 \times 10^{- 10} J$
, Energy in $MeV$
, $E = \frac{1.5 \times 10^{- 10} J}{1.6 \times 10^{- 19}} eV$
, $E = 0.9315 \times 10^{9} eV$
, $E = 931.5 \times 10^{6} eV$
, $E = 931.5 MeV$
, 57. Which one of the following has the highest neutrons ratio?
, (a) $_{92} U^{235}$
, (c) $_{2} {He}^{4}$
, (b) $_{8} O^{16}$
, (d) $_{26} {Fe}^{56}$
]#

1

1.66 \times 10^{- 10} J, 93.15 M e V

2

2.5 \times 10^{- 10} J, 931.5 M e V

3

1.5 \times 10^{- 10} J, 931.5 Me V

4

3 \times 10^{- 10} J, 931.5 M e V

Explanation:

C Ans: a : Neutrons ratio $= \frac{number of neutrons}{number of protons} = \frac{n}{p}$
For $_{92} U^{235}, n = 235 - 92 = 143$
Neutrons ratio $= \frac{143}{92} = 1.55$
For $_{8} O^{16} -$
$n = 16 - 8 = 8$
$p = 8$
Neutrons ratio $= \frac{8}{8} = 1$
For $_{2} {He}^{4} -$
$n = 4 - 2 = 2$
$p = 2$
Neutrons ratio $= \frac{2}{2} = 1$
For $_{26} {Fe}^{56} -$
$n = 56 - 26 = 30$
$p = 26$
Neutrons ratio $= \frac{30}{26} = 1.15$
So, highest neutrons ratio 1.55 for $_{92} U^{235}$ .

NUCLEAR PHYSICS

147402 If elements with principal quantum number $n$ $> 4$ were not allowed in nature, the number of possible elements would have been :

1 32

2 60

3 64

4 4

Explanation:

B We know,
Number of elements $= 2 n^{2}$
Then,
For, $n = 1$
Number of elements $= 2$
For, $n = 2$
Number of elements $= 8$
For, $n = 3$
Number of elements $= 18$
For, $n = 4$
Number of elements $= 32$
So, total number of element $= 32 + 18 + 8 + 2 = 60$

Karnataka CET-2003

NUCLEAR PHYSICS

147394 The number of neutrons in $_{92} U^{235}$ nucleus is

1 327

2 235

3 143

4 92

Explanation:

C Given that,
Mass number $(A) = 235$
Number of protons $(Z) = 92$
Then number of neutrons $(N) = A - Z$
$N = 235 - 92$
$N = 143$

COMEDK 2012

NUCLEAR PHYSICS

147396 In $_{88} {R a}^{226}$ nucleus there are :

1 226 protons and 88 electrons

2 138 protons and 88 neutrons

3 226 neutrons and 138 electrons

4 138 neutrons and 88 protons

Explanation:

D Given that, $_{88} {Ra}^{226}$
Atomic number $(Z) = 88$
Mass number $(A) = 226$
Number of neutrons $= A - Z$
$= 226 - 88$
$= 138$
Number of proton is same as atomic number i.e. 88

AIIMS-1998

NUCLEAR PHYSICS

147397 What is the ratio of the radii of the nuclei of $_{13} {Al}^{27}$ and $_{52} {Te}^{125}$ ?

1

3 : 5

2

1 : 4

3

27 : 125

4

5 : 3

Explanation:

A We know that,
$R = R_{0} A^{1 / 3}$
Then, ratio of radius of $Al$ and $Te -$
$\frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} \Rightarrow \frac{R_{Al}}{R_{Te}} = {(\frac{3^{3}}{5^{3}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = \frac{3}{5}$
$R_{Al} : R_{Te} = 3 : 5$

SCRA-2011

NUCLEAR PHYSICS

147395 Find the energy equivalent of one atomic mass unit in joules and in $M e V$ .
#[Qdiff: Hard, QCat: Numerical Based, examname: AIIMS-2001]
, $1 AMU = 1.6605 \times 10^{- 27} kg, c = 3 \times 10^{8} m / s$
, $E = {mc}^{2}$
, $E = 1.6605 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$
, $E = 1.5 \times 10^{- 10} J$
, Energy in $MeV$
, $E = \frac{1.5 \times 10^{- 10} J}{1.6 \times 10^{- 19}} eV$
, $E = 0.9315 \times 10^{9} eV$
, $E = 931.5 \times 10^{6} eV$
, $E = 931.5 MeV$
, 57. Which one of the following has the highest neutrons ratio?
, (a) $_{92} U^{235}$
, (c) $_{2} {He}^{4}$
, (b) $_{8} O^{16}$
, (d) $_{26} {Fe}^{56}$
]#

1

1.66 \times 10^{- 10} J, 93.15 M e V

2

2.5 \times 10^{- 10} J, 931.5 M e V

3

1.5 \times 10^{- 10} J, 931.5 Me V

4

3 \times 10^{- 10} J, 931.5 M e V

Explanation:

C Ans: a : Neutrons ratio $= \frac{number of neutrons}{number of protons} = \frac{n}{p}$
For $_{92} U^{235}, n = 235 - 92 = 143$
Neutrons ratio $= \frac{143}{92} = 1.55$
For $_{8} O^{16} -$
$n = 16 - 8 = 8$
$p = 8$
Neutrons ratio $= \frac{8}{8} = 1$
For $_{2} {He}^{4} -$
$n = 4 - 2 = 2$
$p = 2$
Neutrons ratio $= \frac{2}{2} = 1$
For $_{26} {Fe}^{56} -$
$n = 56 - 26 = 30$
$p = 26$
Neutrons ratio $= \frac{30}{26} = 1.15$
So, highest neutrons ratio 1.55 for $_{92} U^{235}$ .

NUCLEAR PHYSICS

147402 If elements with principal quantum number $n$ $> 4$ were not allowed in nature, the number of possible elements would have been :

1 32

2 60

3 64

4 4

Explanation:

B We know,
Number of elements $= 2 n^{2}$
Then,
For, $n = 1$
Number of elements $= 2$
For, $n = 2$
Number of elements $= 8$
For, $n = 3$
Number of elements $= 18$
For, $n = 4$
Number of elements $= 32$
So, total number of element $= 32 + 18 + 8 + 2 = 60$

Karnataka CET-2003

NUCLEAR PHYSICS

147394 The number of neutrons in $_{92} U^{235}$ nucleus is

1 327

2 235

3 143

4 92

Explanation:

C Given that,
Mass number $(A) = 235$
Number of protons $(Z) = 92$
Then number of neutrons $(N) = A - Z$
$N = 235 - 92$
$N = 143$

COMEDK 2012

NUCLEAR PHYSICS

147396 In $_{88} {R a}^{226}$ nucleus there are :

1 226 protons and 88 electrons

2 138 protons and 88 neutrons

3 226 neutrons and 138 electrons

4 138 neutrons and 88 protons

Explanation:

D Given that, $_{88} {Ra}^{226}$
Atomic number $(Z) = 88$
Mass number $(A) = 226$
Number of neutrons $= A - Z$
$= 226 - 88$
$= 138$
Number of proton is same as atomic number i.e. 88

AIIMS-1998

NUCLEAR PHYSICS

147397 What is the ratio of the radii of the nuclei of $_{13} {Al}^{27}$ and $_{52} {Te}^{125}$ ?

1

3 : 5

2

1 : 4

3

27 : 125

4

5 : 3

Explanation:

A We know that,
$R = R_{0} A^{1 / 3}$
Then, ratio of radius of $Al$ and $Te -$
$\frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} \Rightarrow \frac{R_{Al}}{R_{Te}} = {(\frac{3^{3}}{5^{3}})}^{1 / 3}$
$\frac{R_{Al}}{R_{Te}} = \frac{3}{5}$
$R_{Al} : R_{Te} = 3 : 5$

SCRA-2011

NUCLEAR PHYSICS

147395 Find the energy equivalent of one atomic mass unit in joules and in $M e V$ .
#[Qdiff: Hard, QCat: Numerical Based, examname: AIIMS-2001]
, $1 AMU = 1.6605 \times 10^{- 27} kg, c = 3 \times 10^{8} m / s$
, $E = {mc}^{2}$
, $E = 1.6605 \times 10^{- 27} \times {(3 \times 10^{8})}^{2}$
, $E = 1.5 \times 10^{- 10} J$
, Energy in $MeV$
, $E = \frac{1.5 \times 10^{- 10} J}{1.6 \times 10^{- 19}} eV$
, $E = 0.9315 \times 10^{9} eV$
, $E = 931.5 \times 10^{6} eV$
, $E = 931.5 MeV$
, 57. Which one of the following has the highest neutrons ratio?
, (a) $_{92} U^{235}$
, (c) $_{2} {He}^{4}$
, (b) $_{8} O^{16}$
, (d) $_{26} {Fe}^{56}$
]#

1

1.66 \times 10^{- 10} J, 93.15 M e V

2

2.5 \times 10^{- 10} J, 931.5 M e V

3

1.5 \times 10^{- 10} J, 931.5 Me V

4

3 \times 10^{- 10} J, 931.5 M e V

Explanation:

C Ans: a : Neutrons ratio $= \frac{number of neutrons}{number of protons} = \frac{n}{p}$
For $_{92} U^{235}, n = 235 - 92 = 143$
Neutrons ratio $= \frac{143}{92} = 1.55$
For $_{8} O^{16} -$
$n = 16 - 8 = 8$
$p = 8$
Neutrons ratio $= \frac{8}{8} = 1$
For $_{2} {He}^{4} -$
$n = 4 - 2 = 2$
$p = 2$
Neutrons ratio $= \frac{2}{2} = 1$
For $_{26} {Fe}^{56} -$
$n = 56 - 26 = 30$
$p = 26$
Neutrons ratio $= \frac{30}{26} = 1.15$
So, highest neutrons ratio 1.55 for $_{92} U^{235}$ .

NUCLEAR PHYSICS

147402 If elements with principal quantum number $n$ $> 4$ were not allowed in nature, the number of possible elements would have been :

1 32

2 60

3 64

4 4

Explanation:

B We know,
Number of elements $= 2 n^{2}$
Then,
For, $n = 1$
Number of elements $= 2$
For, $n = 2$
Number of elements $= 8$
For, $n = 3$
Number of elements $= 18$
For, $n = 4$
Number of elements $= 32$
So, total number of element $= 32 + 18 + 8 + 2 = 60$

Karnataka CET-2003