QBManager

NUCLEAR PHYSICS

147830 A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be

1 1

2

\frac{3}{4}

3

\frac{7}{8}

4

\frac{5}{8}

Explanation:

C Given,
Half life of radio isotope $(T_{1 / 2}) = 5$ years
Total time to decay $(t) = 15$ years
We know that, half lives-
$n = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{o}} = {(\frac{1}{2})}^{3}$
$\frac{N}{N_{o}} = \frac{1}{8}$
The fraction of the atoms decayed will be-
$1 - \frac{1}{8} = \frac{7}{8}$

AMU-2017

NUCLEAR PHYSICS

147831 The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.

1

X

and

Y

have the same decay rate initially

2

X

and

Y

decay at the same rate always

3 Y will decay at a faster rate than

X

4 X will decay at a faster rate than

Y

Explanation:

C We know that,
Rate of decay $= \frac{dN}{dt} = - λ N$
Half-life $(T_{1 / 2}) = \frac{\ln 2}{λ}$
Mean life $(τ) = \frac{1}{λ}$
According to question -
${(T_{1 / 2})}_{X} = (τ)_{Y}$
$\frac{\ln 2}{λ_{X}} = \frac{1}{λ_{Y}}$
$λ_{X} = \ln 2 λ_{Y}$
Hence, $λ_{y} > λ_{x}$ . Initially both of them have the same numbers of atoms the $y$ will decay at a Faster rate than $X$ because $Y$ have higher decay constant than X.

JIPMER-2017

NUCLEAR PHYSICS

147832 Radon-222 has a half-life of 3.8 days. If one starts with $0.064 kg$ of radon-222, the quantity of radon-222 left after 19 days will be

1

0.002 kg

2

0.062 kg

3

0.032 kg

4

0.024 kg

Explanation:

A Given that
Half-life of radon $- 222 (T_{1 / 2}) = 3.8$ days
Initial quantity of radon $(N_{o}) = 0.064 kg$
Total time taken $(t) = 19$ days
Quantity of radioactive substance Radon 222 left,
$N = N_{o} {(\frac{1}{2})}^{n}$
$N = N_{o} {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(∵ n = \frac{t}{T_{1 / 2}})$
$N = 0.064 \times {(\frac{1}{2})}^{\frac{19}{3.8}}$
$N = 0.064 \times {(\frac{1}{2})}^{5}$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 kg$

WB JEE 2017

NUCLEAR PHYSICS

147833 When $_{92} U^{235}$ undergoes fission, about $0.1 %$ of the original mass is converted into energy. The energy released when $1 kg$ of $_{92} U^{235}$ undergoes fission is

1

9 \times 10^{11} J

2

9 \times 10^{13} J

3

9 \times 10^{15} J

4

9 \times 10^{18} J

Explanation:

B According to question-
$Mass = \frac{0.1}{100} \times 1 kg = 10^{- 3} kg$
We know that, $E = Δ {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

J and K CET-2017

NUCLEAR PHYSICS

147830 A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be

1 1

2

\frac{3}{4}

3

\frac{7}{8}

4

\frac{5}{8}

Explanation:

C Given,
Half life of radio isotope $(T_{1 / 2}) = 5$ years
Total time to decay $(t) = 15$ years
We know that, half lives-
$n = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{o}} = {(\frac{1}{2})}^{3}$
$\frac{N}{N_{o}} = \frac{1}{8}$
The fraction of the atoms decayed will be-
$1 - \frac{1}{8} = \frac{7}{8}$

AMU-2017

NUCLEAR PHYSICS

147831 The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.

1

X

and

Y

have the same decay rate initially

2

X

and

Y

decay at the same rate always

3 Y will decay at a faster rate than

X

4 X will decay at a faster rate than

Y

Explanation:

C We know that,
Rate of decay $= \frac{dN}{dt} = - λ N$
Half-life $(T_{1 / 2}) = \frac{\ln 2}{λ}$
Mean life $(τ) = \frac{1}{λ}$
According to question -
${(T_{1 / 2})}_{X} = (τ)_{Y}$
$\frac{\ln 2}{λ_{X}} = \frac{1}{λ_{Y}}$
$λ_{X} = \ln 2 λ_{Y}$
Hence, $λ_{y} > λ_{x}$ . Initially both of them have the same numbers of atoms the $y$ will decay at a Faster rate than $X$ because $Y$ have higher decay constant than X.

JIPMER-2017

NUCLEAR PHYSICS

147832 Radon-222 has a half-life of 3.8 days. If one starts with $0.064 kg$ of radon-222, the quantity of radon-222 left after 19 days will be

1

0.002 kg

2

0.062 kg

3

0.032 kg

4

0.024 kg

Explanation:

A Given that
Half-life of radon $- 222 (T_{1 / 2}) = 3.8$ days
Initial quantity of radon $(N_{o}) = 0.064 kg$
Total time taken $(t) = 19$ days
Quantity of radioactive substance Radon 222 left,
$N = N_{o} {(\frac{1}{2})}^{n}$
$N = N_{o} {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(∵ n = \frac{t}{T_{1 / 2}})$
$N = 0.064 \times {(\frac{1}{2})}^{\frac{19}{3.8}}$
$N = 0.064 \times {(\frac{1}{2})}^{5}$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 kg$

WB JEE 2017

NUCLEAR PHYSICS

147833 When $_{92} U^{235}$ undergoes fission, about $0.1 %$ of the original mass is converted into energy. The energy released when $1 kg$ of $_{92} U^{235}$ undergoes fission is

1

9 \times 10^{11} J

2

9 \times 10^{13} J

3

9 \times 10^{15} J

4

9 \times 10^{18} J

Explanation:

B According to question-
$Mass = \frac{0.1}{100} \times 1 kg = 10^{- 3} kg$
We know that, $E = Δ {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

J and K CET-2017

NUCLEAR PHYSICS

147830 A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be

1 1

2

\frac{3}{4}

3

\frac{7}{8}

4

\frac{5}{8}

Explanation:

C Given,
Half life of radio isotope $(T_{1 / 2}) = 5$ years
Total time to decay $(t) = 15$ years
We know that, half lives-
$n = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{o}} = {(\frac{1}{2})}^{3}$
$\frac{N}{N_{o}} = \frac{1}{8}$
The fraction of the atoms decayed will be-
$1 - \frac{1}{8} = \frac{7}{8}$

AMU-2017

NUCLEAR PHYSICS

147831 The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.

1

X

and

Y

have the same decay rate initially

2

X

and

Y

decay at the same rate always

3 Y will decay at a faster rate than

X

4 X will decay at a faster rate than

Y

Explanation:

C We know that,
Rate of decay $= \frac{dN}{dt} = - λ N$
Half-life $(T_{1 / 2}) = \frac{\ln 2}{λ}$
Mean life $(τ) = \frac{1}{λ}$
According to question -
${(T_{1 / 2})}_{X} = (τ)_{Y}$
$\frac{\ln 2}{λ_{X}} = \frac{1}{λ_{Y}}$
$λ_{X} = \ln 2 λ_{Y}$
Hence, $λ_{y} > λ_{x}$ . Initially both of them have the same numbers of atoms the $y$ will decay at a Faster rate than $X$ because $Y$ have higher decay constant than X.

JIPMER-2017

NUCLEAR PHYSICS

147832 Radon-222 has a half-life of 3.8 days. If one starts with $0.064 kg$ of radon-222, the quantity of radon-222 left after 19 days will be

1

0.002 kg

2

0.062 kg

3

0.032 kg

4

0.024 kg

Explanation:

A Given that
Half-life of radon $- 222 (T_{1 / 2}) = 3.8$ days
Initial quantity of radon $(N_{o}) = 0.064 kg$
Total time taken $(t) = 19$ days
Quantity of radioactive substance Radon 222 left,
$N = N_{o} {(\frac{1}{2})}^{n}$
$N = N_{o} {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(∵ n = \frac{t}{T_{1 / 2}})$
$N = 0.064 \times {(\frac{1}{2})}^{\frac{19}{3.8}}$
$N = 0.064 \times {(\frac{1}{2})}^{5}$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 kg$

WB JEE 2017

NUCLEAR PHYSICS

147833 When $_{92} U^{235}$ undergoes fission, about $0.1 %$ of the original mass is converted into energy. The energy released when $1 kg$ of $_{92} U^{235}$ undergoes fission is

1

9 \times 10^{11} J

2

9 \times 10^{13} J

3

9 \times 10^{15} J

4

9 \times 10^{18} J

Explanation:

B According to question-
$Mass = \frac{0.1}{100} \times 1 kg = 10^{- 3} kg$
We know that, $E = Δ {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

J and K CET-2017

NUCLEAR PHYSICS

147830 A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be

1 1

2

\frac{3}{4}

3

\frac{7}{8}

4

\frac{5}{8}

Explanation:

C Given,
Half life of radio isotope $(T_{1 / 2}) = 5$ years
Total time to decay $(t) = 15$ years
We know that, half lives-
$n = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{o}} = {(\frac{1}{2})}^{3}$
$\frac{N}{N_{o}} = \frac{1}{8}$
The fraction of the atoms decayed will be-
$1 - \frac{1}{8} = \frac{7}{8}$

AMU-2017

NUCLEAR PHYSICS

147831 The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.

1

X

and

Y

have the same decay rate initially

2

X

and

Y

decay at the same rate always

3 Y will decay at a faster rate than

X

4 X will decay at a faster rate than

Y

Explanation:

C We know that,
Rate of decay $= \frac{dN}{dt} = - λ N$
Half-life $(T_{1 / 2}) = \frac{\ln 2}{λ}$
Mean life $(τ) = \frac{1}{λ}$
According to question -
${(T_{1 / 2})}_{X} = (τ)_{Y}$
$\frac{\ln 2}{λ_{X}} = \frac{1}{λ_{Y}}$
$λ_{X} = \ln 2 λ_{Y}$
Hence, $λ_{y} > λ_{x}$ . Initially both of them have the same numbers of atoms the $y$ will decay at a Faster rate than $X$ because $Y$ have higher decay constant than X.

JIPMER-2017

NUCLEAR PHYSICS

147832 Radon-222 has a half-life of 3.8 days. If one starts with $0.064 kg$ of radon-222, the quantity of radon-222 left after 19 days will be

1

0.002 kg

2

0.062 kg

3

0.032 kg

4

0.024 kg

Explanation:

A Given that
Half-life of radon $- 222 (T_{1 / 2}) = 3.8$ days
Initial quantity of radon $(N_{o}) = 0.064 kg$
Total time taken $(t) = 19$ days
Quantity of radioactive substance Radon 222 left,
$N = N_{o} {(\frac{1}{2})}^{n}$
$N = N_{o} {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(∵ n = \frac{t}{T_{1 / 2}})$
$N = 0.064 \times {(\frac{1}{2})}^{\frac{19}{3.8}}$
$N = 0.064 \times {(\frac{1}{2})}^{5}$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 kg$

WB JEE 2017

NUCLEAR PHYSICS

147833 When $_{92} U^{235}$ undergoes fission, about $0.1 %$ of the original mass is converted into energy. The energy released when $1 kg$ of $_{92} U^{235}$ undergoes fission is

1

9 \times 10^{11} J

2

9 \times 10^{13} J

3

9 \times 10^{15} J

4

9 \times 10^{18} J

Explanation:

B According to question-
$Mass = \frac{0.1}{100} \times 1 kg = 10^{- 3} kg$
We know that, $E = Δ {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

J and K CET-2017