QBManager

NUCLEAR PHYSICS

147562 If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \min$ . will be

1

\frac{1}{8}

2

\frac{1}{2}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

A Given,
Half life $T_{1 / 2} = 30 \min$ . radioactivity time $T = 90$ minute.
Then, $n = \frac{T}{T_{1 / 2}} = \frac{90}{30} = 3$
Initial amount $= N_{o}$
Undecayed amount,
$N = N_{o} \times {(\frac{1}{2})}^{n}$
$N = N_{o} \times {(\frac{1}{2})}^{3}$
$N = \frac{N_{o}}{8}$
$\frac{N}{N_{o}} = \frac{1}{8}$

JEE Main-29.01.2023

NUCLEAR PHYSICS

147564 A nuclear reactor delivers a power of $10^{9} W$ , the amount of fuel consumed by the reactor in one hour is

1

0.72 g

2

0.96 g

3

0.04 g

4

0.08 g

Explanation:

C Given,
Power $(P) = 10^{9} W$
Power $(P) = \frac{E}{t} = \frac{{mc}^{2}}{t}$
Mass $(m) = \frac{Pt}{c^{2}}$
$= \frac{10^{9} \times 60 \times 60}{{(3 \times 10^{8})}^{2}}$
$= \frac{36 \times 10^{11}}{9 \times 10^{16}}$
$= 4 \times 10^{- 5} kg$
$= 0.04 g$

Karnataka CET-2022

NUCLEAR PHYSICS

147565 Calculate the energy released during the alpha decay of $_{92}^{238} U = 238$ . Given the following data.
$_{92}^{238} U = 238.05079 u,_{90}^{234} Th = 234.04363 u$
$_{2}^{4} He = 4.00260 u, 1 u = 931.5 MeV / C^{2}$ .

1

4.25 MeV

2

3.75 MeV

3

3.50 MeV

4

2.75 MeV

Explanation:

A Given,
Mass of $_{92}^{298} U = 238.05079 u$
Mass of $_{90}^{234} Th = 234.04363 u$
$∴$ Mass defect $=$ mass of uranium - mass of thorium mass of helium
$= 238.05079 u - 234.04363 u - 4.0026 u$
$= 0.0045644 u$
Now, $1 u$ release $931.5 MeV / C^{2}$ energy.
Energy released by $0.00456 = (0.00456 \times 931.5) \frac{MeV}{C^{2}}$
$⊔ 4.25 MeV$

AMU-2019

NUCLEAR PHYSICS

147566 A radioactive element $A$ converts into another stable element B. Half life of $A$ is $1.5$ hrs. After time $t$ the ratio of atoms of $A$ and $B$ is found to be $1 : 8$ , then $t$ in hours is-

1 6

2 8

3 Between 3 to 4.5

4 Between 4.5 to 6

Explanation:

D Let $N_{o}$ be the number of atoms of $x$ at time $t$ $= 0$ then, at (half life)
So, $N_{x} = \frac{N_{0}}{8}$ and $N_{y} = N_{0} - \frac{N_{0}}{8} = \frac{7 N_{0}}{8}$
$\frac{N_{x}}{N_{y}} = \frac{1}{7}$
And at, $t = 6 h$ (4 half life)
$N_{x} = \frac{N_{0}}{16}$ and $N_{y} = \frac{15 N_{0}}{16}$
$\frac{N_{x}}{N_{y}} = \frac{1}{15}$
The given ratio $\frac{1}{8}$ lies between $\frac{1}{7}$ and $\frac{1}{15}$
Therefore, $t$ lies between $4.5 h$ to $6 h$

AP EAMCET-08.07.2022

NUCLEAR PHYSICS

147562 If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \min$ . will be

1

\frac{1}{8}

2

\frac{1}{2}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

A Given,
Half life $T_{1 / 2} = 30 \min$ . radioactivity time $T = 90$ minute.
Then, $n = \frac{T}{T_{1 / 2}} = \frac{90}{30} = 3$
Initial amount $= N_{o}$
Undecayed amount,
$N = N_{o} \times {(\frac{1}{2})}^{n}$
$N = N_{o} \times {(\frac{1}{2})}^{3}$
$N = \frac{N_{o}}{8}$
$\frac{N}{N_{o}} = \frac{1}{8}$

JEE Main-29.01.2023

NUCLEAR PHYSICS

147564 A nuclear reactor delivers a power of $10^{9} W$ , the amount of fuel consumed by the reactor in one hour is

1

0.72 g

2

0.96 g

3

0.04 g

4

0.08 g

Explanation:

C Given,
Power $(P) = 10^{9} W$
Power $(P) = \frac{E}{t} = \frac{{mc}^{2}}{t}$
Mass $(m) = \frac{Pt}{c^{2}}$
$= \frac{10^{9} \times 60 \times 60}{{(3 \times 10^{8})}^{2}}$
$= \frac{36 \times 10^{11}}{9 \times 10^{16}}$
$= 4 \times 10^{- 5} kg$
$= 0.04 g$

Karnataka CET-2022

NUCLEAR PHYSICS

147565 Calculate the energy released during the alpha decay of $_{92}^{238} U = 238$ . Given the following data.
$_{92}^{238} U = 238.05079 u,_{90}^{234} Th = 234.04363 u$
$_{2}^{4} He = 4.00260 u, 1 u = 931.5 MeV / C^{2}$ .

1

4.25 MeV

2

3.75 MeV

3

3.50 MeV

4

2.75 MeV

Explanation:

A Given,
Mass of $_{92}^{298} U = 238.05079 u$
Mass of $_{90}^{234} Th = 234.04363 u$
$∴$ Mass defect $=$ mass of uranium - mass of thorium mass of helium
$= 238.05079 u - 234.04363 u - 4.0026 u$
$= 0.0045644 u$
Now, $1 u$ release $931.5 MeV / C^{2}$ energy.
Energy released by $0.00456 = (0.00456 \times 931.5) \frac{MeV}{C^{2}}$
$⊔ 4.25 MeV$

AMU-2019

NUCLEAR PHYSICS

147566 A radioactive element $A$ converts into another stable element B. Half life of $A$ is $1.5$ hrs. After time $t$ the ratio of atoms of $A$ and $B$ is found to be $1 : 8$ , then $t$ in hours is-

1 6

2 8

3 Between 3 to 4.5

4 Between 4.5 to 6

Explanation:

D Let $N_{o}$ be the number of atoms of $x$ at time $t$ $= 0$ then, at (half life)
So, $N_{x} = \frac{N_{0}}{8}$ and $N_{y} = N_{0} - \frac{N_{0}}{8} = \frac{7 N_{0}}{8}$
$\frac{N_{x}}{N_{y}} = \frac{1}{7}$
And at, $t = 6 h$ (4 half life)
$N_{x} = \frac{N_{0}}{16}$ and $N_{y} = \frac{15 N_{0}}{16}$
$\frac{N_{x}}{N_{y}} = \frac{1}{15}$
The given ratio $\frac{1}{8}$ lies between $\frac{1}{7}$ and $\frac{1}{15}$
Therefore, $t$ lies between $4.5 h$ to $6 h$

AP EAMCET-08.07.2022

NUCLEAR PHYSICS

147562 If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \min$ . will be

1

\frac{1}{8}

2

\frac{1}{2}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

A Given,
Half life $T_{1 / 2} = 30 \min$ . radioactivity time $T = 90$ minute.
Then, $n = \frac{T}{T_{1 / 2}} = \frac{90}{30} = 3$
Initial amount $= N_{o}$
Undecayed amount,
$N = N_{o} \times {(\frac{1}{2})}^{n}$
$N = N_{o} \times {(\frac{1}{2})}^{3}$
$N = \frac{N_{o}}{8}$
$\frac{N}{N_{o}} = \frac{1}{8}$

JEE Main-29.01.2023

NUCLEAR PHYSICS

147564 A nuclear reactor delivers a power of $10^{9} W$ , the amount of fuel consumed by the reactor in one hour is

1

0.72 g

2

0.96 g

3

0.04 g

4

0.08 g

Explanation:

C Given,
Power $(P) = 10^{9} W$
Power $(P) = \frac{E}{t} = \frac{{mc}^{2}}{t}$
Mass $(m) = \frac{Pt}{c^{2}}$
$= \frac{10^{9} \times 60 \times 60}{{(3 \times 10^{8})}^{2}}$
$= \frac{36 \times 10^{11}}{9 \times 10^{16}}$
$= 4 \times 10^{- 5} kg$
$= 0.04 g$

Karnataka CET-2022

NUCLEAR PHYSICS

147565 Calculate the energy released during the alpha decay of $_{92}^{238} U = 238$ . Given the following data.
$_{92}^{238} U = 238.05079 u,_{90}^{234} Th = 234.04363 u$
$_{2}^{4} He = 4.00260 u, 1 u = 931.5 MeV / C^{2}$ .

1

4.25 MeV

2

3.75 MeV

3

3.50 MeV

4

2.75 MeV

Explanation:

A Given,
Mass of $_{92}^{298} U = 238.05079 u$
Mass of $_{90}^{234} Th = 234.04363 u$
$∴$ Mass defect $=$ mass of uranium - mass of thorium mass of helium
$= 238.05079 u - 234.04363 u - 4.0026 u$
$= 0.0045644 u$
Now, $1 u$ release $931.5 MeV / C^{2}$ energy.
Energy released by $0.00456 = (0.00456 \times 931.5) \frac{MeV}{C^{2}}$
$⊔ 4.25 MeV$

AMU-2019

NUCLEAR PHYSICS

147566 A radioactive element $A$ converts into another stable element B. Half life of $A$ is $1.5$ hrs. After time $t$ the ratio of atoms of $A$ and $B$ is found to be $1 : 8$ , then $t$ in hours is-

1 6

2 8

3 Between 3 to 4.5

4 Between 4.5 to 6

Explanation:

D Let $N_{o}$ be the number of atoms of $x$ at time $t$ $= 0$ then, at (half life)
So, $N_{x} = \frac{N_{0}}{8}$ and $N_{y} = N_{0} - \frac{N_{0}}{8} = \frac{7 N_{0}}{8}$
$\frac{N_{x}}{N_{y}} = \frac{1}{7}$
And at, $t = 6 h$ (4 half life)
$N_{x} = \frac{N_{0}}{16}$ and $N_{y} = \frac{15 N_{0}}{16}$
$\frac{N_{x}}{N_{y}} = \frac{1}{15}$
The given ratio $\frac{1}{8}$ lies between $\frac{1}{7}$ and $\frac{1}{15}$
Therefore, $t$ lies between $4.5 h$ to $6 h$

AP EAMCET-08.07.2022

NUCLEAR PHYSICS

147562 If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \min$ . will be

1

\frac{1}{8}

2

\frac{1}{2}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

A Given,
Half life $T_{1 / 2} = 30 \min$ . radioactivity time $T = 90$ minute.
Then, $n = \frac{T}{T_{1 / 2}} = \frac{90}{30} = 3$
Initial amount $= N_{o}$
Undecayed amount,
$N = N_{o} \times {(\frac{1}{2})}^{n}$
$N = N_{o} \times {(\frac{1}{2})}^{3}$
$N = \frac{N_{o}}{8}$
$\frac{N}{N_{o}} = \frac{1}{8}$

JEE Main-29.01.2023

NUCLEAR PHYSICS

147564 A nuclear reactor delivers a power of $10^{9} W$ , the amount of fuel consumed by the reactor in one hour is

1

0.72 g

2

0.96 g

3

0.04 g

4

0.08 g

Explanation:

C Given,
Power $(P) = 10^{9} W$
Power $(P) = \frac{E}{t} = \frac{{mc}^{2}}{t}$
Mass $(m) = \frac{Pt}{c^{2}}$
$= \frac{10^{9} \times 60 \times 60}{{(3 \times 10^{8})}^{2}}$
$= \frac{36 \times 10^{11}}{9 \times 10^{16}}$
$= 4 \times 10^{- 5} kg$
$= 0.04 g$

Karnataka CET-2022

NUCLEAR PHYSICS

147565 Calculate the energy released during the alpha decay of $_{92}^{238} U = 238$ . Given the following data.
$_{92}^{238} U = 238.05079 u,_{90}^{234} Th = 234.04363 u$
$_{2}^{4} He = 4.00260 u, 1 u = 931.5 MeV / C^{2}$ .

1

4.25 MeV

2

3.75 MeV

3

3.50 MeV

4

2.75 MeV

Explanation:

A Given,
Mass of $_{92}^{298} U = 238.05079 u$
Mass of $_{90}^{234} Th = 234.04363 u$
$∴$ Mass defect $=$ mass of uranium - mass of thorium mass of helium
$= 238.05079 u - 234.04363 u - 4.0026 u$
$= 0.0045644 u$
Now, $1 u$ release $931.5 MeV / C^{2}$ energy.
Energy released by $0.00456 = (0.00456 \times 931.5) \frac{MeV}{C^{2}}$
$⊔ 4.25 MeV$

AMU-2019

NUCLEAR PHYSICS

147566 A radioactive element $A$ converts into another stable element B. Half life of $A$ is $1.5$ hrs. After time $t$ the ratio of atoms of $A$ and $B$ is found to be $1 : 8$ , then $t$ in hours is-

1 6

2 8

3 Between 3 to 4.5

4 Between 4.5 to 6

Explanation:

D Let $N_{o}$ be the number of atoms of $x$ at time $t$ $= 0$ then, at (half life)
So, $N_{x} = \frac{N_{0}}{8}$ and $N_{y} = N_{0} - \frac{N_{0}}{8} = \frac{7 N_{0}}{8}$
$\frac{N_{x}}{N_{y}} = \frac{1}{7}$
And at, $t = 6 h$ (4 half life)
$N_{x} = \frac{N_{0}}{16}$ and $N_{y} = \frac{15 N_{0}}{16}$
$\frac{N_{x}}{N_{y}} = \frac{1}{15}$
The given ratio $\frac{1}{8}$ lies between $\frac{1}{7}$ and $\frac{1}{15}$
Therefore, $t$ lies between $4.5 h$ to $6 h$

AP EAMCET-08.07.2022

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: