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NUCLEAR PHYSICS

147968 If the energy released in the fission of one nucleus is $3.2 \times 10^{- 11} J$ . Then, the numbers of nuclei required per second in a power plant of $16 kW$ is (assume efficiency of plant $= 1 %$ )
#[Qdiff: Hard, QCat: Numerical Based, examname: UP CPMT-2001,AP EAMCET- 24 -09-2020] Shift - I
]#

1

5 \times 10^{12}

2

5 \times 10^{14}

3

5 \times 10^{16}

4

5.12 \times 10^{16}

Explanation:

B Given that,
Energy released $(E) = 3.2 \times 10^{- 11} J$
Power $(P) = 16 kW$
$= 16 \times 10^{3} watt$
Number of nuclei required per second $= \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}}$
$= \frac{16}{3.2} \times 10^{3} \times 10^{11}$
$= 5 \times 10^{14}$

NUCLEAR PHYSICS

147969 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{A I}$ then the radius of $_{53}^{125} Te$ nucleus is

1

\frac{5}{3} R_{A l}

2

\frac{13}{53} R_{Al}

3

\frac{3}{5} R_{A l}

4

\frac{53}{13} R_{A l}

Explanation:

A We know that,
$R = R_{0} (A)^{1 / 3}$
So, $R_{Al} = R_{0} (27)^{1 / 3}$
$R_{A l} = 3 R_{0}$
$R_{0} = \frac{R_{A l}}{3}$
And $R_{Te} = R_{0} (125) V_{3}$
$R_{Te} = R_{0} \times 5$
$R_{0} = \frac{R_{Te}}{5}$
On comparing equation (i) and (ii) we get -
$\frac{R_{Te}}{5} = \frac{R_{Al}}{3}$
$R_{Te} = \frac{5}{3} R_{A l}$

MHT-CET 2020

NUCLEAR PHYSICS

147972 The energy released when one nucleus of $_{92} U^{235}$ undergoes fission is $188 MeV$ . The energy releases when $100 g$ of $_{92} U^{235}$ undergoes fission is

1

3.55 \times 10^{12} J

2

7.71 \times 10^{12} J

3

3.55 \times 10^{13} J

4

7.71 \times 10^{13} J

Explanation:

B Each Nucleus, $E_{0} = 188 \times 10^{6} \times 1.6 \times 10^{- 19} J$
For, $_{92} U^{235}$
$E_{0} = 300.8 \times 10^{- 13} J$
235 gms $\to 6 \times 10^{23}$ Nuclei
100 gms $\to N$
$\frac{235}{100} = \frac{6 \times 10^{23}}{N}$
$N = \frac{100}{235} \times 6 \times 10^{23}$
$N = 2.5532 \times 10^{23}$ Nuclei
Total energy,
$T . E = E_{o} \times N$
T.E $= 300.8 \times 10^{- 13} \times 2.5532 \times 10^{23}$
$T . E = 771 \times 10^{10} = 768.002 \times 10^{10}$
T.E $= 7.71 \times 10^{12} J o u l e = 7.71 \times 10^{12}$ Joule.

AP EAMCET (23.04.2019) Shift-I

NUCLEAR PHYSICS

147976 In uranium radioactive series, initial nucleus $^{238} U_{92}$ decays to final nucleus $^{206} U_{82}$ . In this process, the number of $α$ -particles and $β$ particles emitted are

1 8 and 3

2 16 and 6

3 16 and 3

4 8 and 6

Explanation:

D Chemical reaction-
$^{238} U_{92} \to^{206} U_{82} + x α_{2}^{4} + y β_{- 1}^{0}$
Let $x$ and $y$ be the number of $α$ and $β$ particle respectively.
Applying law of conservation of atomic number, we get-
$92 = 82 + 2 x - y$
$2 x - y = 10$
On applying law of conservation of mass we get-
$238 = 206 + 4 x$
$4 x = 32$
$x = 8$
On putting the value of $x$ in equation (i), we get-
$2 x - y = 10$
$2 \times 8 - y = 10$
$- y = 10 - 16$
$- y = - 6$
$y = 6$
So, 8 and $6, α$ and $β$ -particles are emitted.

TS- EAMCET-05.05.2018

NUCLEAR PHYSICS

147968 If the energy released in the fission of one nucleus is $3.2 \times 10^{- 11} J$ . Then, the numbers of nuclei required per second in a power plant of $16 kW$ is (assume efficiency of plant $= 1 %$ )
#[Qdiff: Hard, QCat: Numerical Based, examname: UP CPMT-2001,AP EAMCET- 24 -09-2020] Shift - I
]#

1

5 \times 10^{12}

2

5 \times 10^{14}

3

5 \times 10^{16}

4

5.12 \times 10^{16}

Explanation:

B Given that,
Energy released $(E) = 3.2 \times 10^{- 11} J$
Power $(P) = 16 kW$
$= 16 \times 10^{3} watt$
Number of nuclei required per second $= \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}}$
$= \frac{16}{3.2} \times 10^{3} \times 10^{11}$
$= 5 \times 10^{14}$

NUCLEAR PHYSICS

147969 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{A I}$ then the radius of $_{53}^{125} Te$ nucleus is

1

\frac{5}{3} R_{A l}

2

\frac{13}{53} R_{Al}

3

\frac{3}{5} R_{A l}

4

\frac{53}{13} R_{A l}

Explanation:

A We know that,
$R = R_{0} (A)^{1 / 3}$
So, $R_{Al} = R_{0} (27)^{1 / 3}$
$R_{A l} = 3 R_{0}$
$R_{0} = \frac{R_{A l}}{3}$
And $R_{Te} = R_{0} (125) V_{3}$
$R_{Te} = R_{0} \times 5$
$R_{0} = \frac{R_{Te}}{5}$
On comparing equation (i) and (ii) we get -
$\frac{R_{Te}}{5} = \frac{R_{Al}}{3}$
$R_{Te} = \frac{5}{3} R_{A l}$

MHT-CET 2020

NUCLEAR PHYSICS

147972 The energy released when one nucleus of $_{92} U^{235}$ undergoes fission is $188 MeV$ . The energy releases when $100 g$ of $_{92} U^{235}$ undergoes fission is

1

3.55 \times 10^{12} J

2

7.71 \times 10^{12} J

3

3.55 \times 10^{13} J

4

7.71 \times 10^{13} J

Explanation:

B Each Nucleus, $E_{0} = 188 \times 10^{6} \times 1.6 \times 10^{- 19} J$
For, $_{92} U^{235}$
$E_{0} = 300.8 \times 10^{- 13} J$
235 gms $\to 6 \times 10^{23}$ Nuclei
100 gms $\to N$
$\frac{235}{100} = \frac{6 \times 10^{23}}{N}$
$N = \frac{100}{235} \times 6 \times 10^{23}$
$N = 2.5532 \times 10^{23}$ Nuclei
Total energy,
$T . E = E_{o} \times N$
T.E $= 300.8 \times 10^{- 13} \times 2.5532 \times 10^{23}$
$T . E = 771 \times 10^{10} = 768.002 \times 10^{10}$
T.E $= 7.71 \times 10^{12} J o u l e = 7.71 \times 10^{12}$ Joule.

AP EAMCET (23.04.2019) Shift-I

NUCLEAR PHYSICS

147976 In uranium radioactive series, initial nucleus $^{238} U_{92}$ decays to final nucleus $^{206} U_{82}$ . In this process, the number of $α$ -particles and $β$ particles emitted are

1 8 and 3

2 16 and 6

3 16 and 3

4 8 and 6

Explanation:

D Chemical reaction-
$^{238} U_{92} \to^{206} U_{82} + x α_{2}^{4} + y β_{- 1}^{0}$
Let $x$ and $y$ be the number of $α$ and $β$ particle respectively.
Applying law of conservation of atomic number, we get-
$92 = 82 + 2 x - y$
$2 x - y = 10$
On applying law of conservation of mass we get-
$238 = 206 + 4 x$
$4 x = 32$
$x = 8$
On putting the value of $x$ in equation (i), we get-
$2 x - y = 10$
$2 \times 8 - y = 10$
$- y = 10 - 16$
$- y = - 6$
$y = 6$
So, 8 and $6, α$ and $β$ -particles are emitted.

TS- EAMCET-05.05.2018

NUCLEAR PHYSICS

147968 If the energy released in the fission of one nucleus is $3.2 \times 10^{- 11} J$ . Then, the numbers of nuclei required per second in a power plant of $16 kW$ is (assume efficiency of plant $= 1 %$ )
#[Qdiff: Hard, QCat: Numerical Based, examname: UP CPMT-2001,AP EAMCET- 24 -09-2020] Shift - I
]#

1

5 \times 10^{12}

2

5 \times 10^{14}

3

5 \times 10^{16}

4

5.12 \times 10^{16}

Explanation:

B Given that,
Energy released $(E) = 3.2 \times 10^{- 11} J$
Power $(P) = 16 kW$
$= 16 \times 10^{3} watt$
Number of nuclei required per second $= \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}}$
$= \frac{16}{3.2} \times 10^{3} \times 10^{11}$
$= 5 \times 10^{14}$

NUCLEAR PHYSICS

147969 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{A I}$ then the radius of $_{53}^{125} Te$ nucleus is

1

\frac{5}{3} R_{A l}

2

\frac{13}{53} R_{Al}

3

\frac{3}{5} R_{A l}

4

\frac{53}{13} R_{A l}

Explanation:

A We know that,
$R = R_{0} (A)^{1 / 3}$
So, $R_{Al} = R_{0} (27)^{1 / 3}$
$R_{A l} = 3 R_{0}$
$R_{0} = \frac{R_{A l}}{3}$
And $R_{Te} = R_{0} (125) V_{3}$
$R_{Te} = R_{0} \times 5$
$R_{0} = \frac{R_{Te}}{5}$
On comparing equation (i) and (ii) we get -
$\frac{R_{Te}}{5} = \frac{R_{Al}}{3}$
$R_{Te} = \frac{5}{3} R_{A l}$

MHT-CET 2020

NUCLEAR PHYSICS

147972 The energy released when one nucleus of $_{92} U^{235}$ undergoes fission is $188 MeV$ . The energy releases when $100 g$ of $_{92} U^{235}$ undergoes fission is

1

3.55 \times 10^{12} J

2

7.71 \times 10^{12} J

3

3.55 \times 10^{13} J

4

7.71 \times 10^{13} J

Explanation:

B Each Nucleus, $E_{0} = 188 \times 10^{6} \times 1.6 \times 10^{- 19} J$
For, $_{92} U^{235}$
$E_{0} = 300.8 \times 10^{- 13} J$
235 gms $\to 6 \times 10^{23}$ Nuclei
100 gms $\to N$
$\frac{235}{100} = \frac{6 \times 10^{23}}{N}$
$N = \frac{100}{235} \times 6 \times 10^{23}$
$N = 2.5532 \times 10^{23}$ Nuclei
Total energy,
$T . E = E_{o} \times N$
T.E $= 300.8 \times 10^{- 13} \times 2.5532 \times 10^{23}$
$T . E = 771 \times 10^{10} = 768.002 \times 10^{10}$
T.E $= 7.71 \times 10^{12} J o u l e = 7.71 \times 10^{12}$ Joule.

AP EAMCET (23.04.2019) Shift-I

NUCLEAR PHYSICS

147976 In uranium radioactive series, initial nucleus $^{238} U_{92}$ decays to final nucleus $^{206} U_{82}$ . In this process, the number of $α$ -particles and $β$ particles emitted are

1 8 and 3

2 16 and 6

3 16 and 3

4 8 and 6

Explanation:

D Chemical reaction-
$^{238} U_{92} \to^{206} U_{82} + x α_{2}^{4} + y β_{- 1}^{0}$
Let $x$ and $y$ be the number of $α$ and $β$ particle respectively.
Applying law of conservation of atomic number, we get-
$92 = 82 + 2 x - y$
$2 x - y = 10$
On applying law of conservation of mass we get-
$238 = 206 + 4 x$
$4 x = 32$
$x = 8$
On putting the value of $x$ in equation (i), we get-
$2 x - y = 10$
$2 \times 8 - y = 10$
$- y = 10 - 16$
$- y = - 6$
$y = 6$
So, 8 and $6, α$ and $β$ -particles are emitted.

TS- EAMCET-05.05.2018

NUCLEAR PHYSICS

147968 If the energy released in the fission of one nucleus is $3.2 \times 10^{- 11} J$ . Then, the numbers of nuclei required per second in a power plant of $16 kW$ is (assume efficiency of plant $= 1 %$ )
#[Qdiff: Hard, QCat: Numerical Based, examname: UP CPMT-2001,AP EAMCET- 24 -09-2020] Shift - I
]#

1

5 \times 10^{12}

2

5 \times 10^{14}

3

5 \times 10^{16}

4

5.12 \times 10^{16}

Explanation:

B Given that,
Energy released $(E) = 3.2 \times 10^{- 11} J$
Power $(P) = 16 kW$
$= 16 \times 10^{3} watt$
Number of nuclei required per second $= \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}}$
$= \frac{16}{3.2} \times 10^{3} \times 10^{11}$
$= 5 \times 10^{14}$

NUCLEAR PHYSICS

147969 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{A I}$ then the radius of $_{53}^{125} Te$ nucleus is

1

\frac{5}{3} R_{A l}

2

\frac{13}{53} R_{Al}

3

\frac{3}{5} R_{A l}

4

\frac{53}{13} R_{A l}

Explanation:

A We know that,
$R = R_{0} (A)^{1 / 3}$
So, $R_{Al} = R_{0} (27)^{1 / 3}$
$R_{A l} = 3 R_{0}$
$R_{0} = \frac{R_{A l}}{3}$
And $R_{Te} = R_{0} (125) V_{3}$
$R_{Te} = R_{0} \times 5$
$R_{0} = \frac{R_{Te}}{5}$
On comparing equation (i) and (ii) we get -
$\frac{R_{Te}}{5} = \frac{R_{Al}}{3}$
$R_{Te} = \frac{5}{3} R_{A l}$

MHT-CET 2020

NUCLEAR PHYSICS

147972 The energy released when one nucleus of $_{92} U^{235}$ undergoes fission is $188 MeV$ . The energy releases when $100 g$ of $_{92} U^{235}$ undergoes fission is

1

3.55 \times 10^{12} J

2

7.71 \times 10^{12} J

3

3.55 \times 10^{13} J

4

7.71 \times 10^{13} J

Explanation:

B Each Nucleus, $E_{0} = 188 \times 10^{6} \times 1.6 \times 10^{- 19} J$
For, $_{92} U^{235}$
$E_{0} = 300.8 \times 10^{- 13} J$
235 gms $\to 6 \times 10^{23}$ Nuclei
100 gms $\to N$
$\frac{235}{100} = \frac{6 \times 10^{23}}{N}$
$N = \frac{100}{235} \times 6 \times 10^{23}$
$N = 2.5532 \times 10^{23}$ Nuclei
Total energy,
$T . E = E_{o} \times N$
T.E $= 300.8 \times 10^{- 13} \times 2.5532 \times 10^{23}$
$T . E = 771 \times 10^{10} = 768.002 \times 10^{10}$
T.E $= 7.71 \times 10^{12} J o u l e = 7.71 \times 10^{12}$ Joule.

AP EAMCET (23.04.2019) Shift-I

NUCLEAR PHYSICS

147976 In uranium radioactive series, initial nucleus $^{238} U_{92}$ decays to final nucleus $^{206} U_{82}$ . In this process, the number of $α$ -particles and $β$ particles emitted are

1 8 and 3

2 16 and 6

3 16 and 3

4 8 and 6

Explanation:

D Chemical reaction-
$^{238} U_{92} \to^{206} U_{82} + x α_{2}^{4} + y β_{- 1}^{0}$
Let $x$ and $y$ be the number of $α$ and $β$ particle respectively.
Applying law of conservation of atomic number, we get-
$92 = 82 + 2 x - y$
$2 x - y = 10$
On applying law of conservation of mass we get-
$238 = 206 + 4 x$
$4 x = 32$
$x = 8$
On putting the value of $x$ in equation (i), we get-
$2 x - y = 10$
$2 \times 8 - y = 10$
$- y = 10 - 16$
$- y = - 6$
$y = 6$
So, 8 and $6, α$ and $β$ -particles are emitted.

TS- EAMCET-05.05.2018