QBManager

NUCLEAR PHYSICS

147581 In the Uranium radioactive series, the initial nucleus is $_{92}^{238} U$ and final nucleus is $_{82}^{206} Pb$ . When the Uranium nucleus decays to lead, the number of $α$ -particles emitted is ........ and the number of $β$ -particles emitted is

1 6,8

2 8,6

3 16,6

4 32,2

Explanation:

B When an a particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4
The number of alpha particles emitted is $\frac{238 - 206}{4} = \frac{32}{4} = 8$
When 8 alpha particles are emitted the atomic number decreases by $2 \times 8 = 16$ from 92 to $92 - 16 = 76$
When $β$ -particles is emitted the atomic number increases by 1 .
The number of $β$ -particles emitted $= 82 - 76 = 6$ .
Hence, the number of $α$ -particle emitted is 8 and the number of $β$ -particle emitted is 6 .

AP EAMCET (18.09.2020) Shift-II

NUCLEAR PHYSICS

147583 A radioactive source has a half-life of $6 h$ . A freshly prepared sample of the same exhibits radioactivity 32 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is

1

30 h

2

24 h

3

18 h

4

12 h

Explanation:

A Let's say the permissible level is N, so initially we have $32 N$ nuclei present.
We need to calculate the time taken to drop the number of nuclei from $32 N$ to $N$ .
So, $N = N_{0} e^{- λ t}$
$N_{o}$ is $32 N$ so after substituting this we get,
$e^{- λ t} = \frac{1}{32}$
$e^{λ t} = 32$
$λ t = \ln 32$
$t = \frac{\ln 32}{λ}$
We have, half life $(t_{1 / 2}) = 6 h$
$λ = \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 2}{6}$
Putting value of equation (ii) in equation (i), we get-
$t = \frac{\ln 32}{\ln 2} \times 6$
$t = \frac{5 \ln 2}{\ln 2} \times 6$
$t = 30 hours$

TS- EAMCET-09.09.2020

NUCLEAR PHYSICS

147584 The half-life of a radioactive isotope is $30 h$ . How long will it take to get reduced to $12.5 %$ of its initial amount?

1

120 h

2

90 h

3

60 h

4

50 h

Explanation:

B Given that,
Half life of radioactive sample $(T_{1 / 2}) = 30 h$
After decay, the remaining amount is $N$ -
So, $N = 12.5 %$ of $N_{0}$
$N = \frac{125}{1000} \times N_{0}$
$N = \frac{N_{0}}{8}$
As we know that,
$N = N_{0} {(\frac{1}{2})}^{n}$
From the equation (i) and equation (ii) we have-
$\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{n}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n}$
$n = 3$
Then,
$\frac{t}{T_{1 / 2}} = 3$
$\frac{t}{30} = 3$
$t = 90 h$

TS- EAMCET.14.09.2020

NUCLEAR PHYSICS

147585 Half-life of radioactive sample is $24 h$ . If a newly prepared radioactive sample shows 4 times the allowed and safe value of radio activity, the minimum time after which one can work safety with the source is

1

48 h

2

96 h

3

8 h

4

72 h

Explanation:

A Given, half life $(T_{1 / 2}) = 24 h$
According to question -
$\frac{N_{0}}{4 N_{o}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{t}{T_{1 / 2}} = 2$
$t = 2 \times T_{1 / 2}$
$t = 2 \times 24$
$t = 48 h$

TS- EAMCET-14.09.2020

NUCLEAR PHYSICS

147586 The half-life of a radioactive sample is $5 s$ . If the initial mass of the sample is $60 g$ , then the time required to reduce the sample to $7.5 g$ is

1

15 s

2

75 s

3

7.5 s

4

10 s

Explanation:

A Given,
Half life $(t_{1 / 2}) = 5 s$
Initial mass of sample $(N_{o}) = 60 g$
Reduce mass of sample $(N) = 7.5 g$
As we know,
$Half life (t_{1 / 2}) = \frac{\ln 2}{λ}$
$λ = \frac{0.693}{5} = 0.1386$
According to radioactivity decay law-
$N = N_{o} e^{- λ t}$
$7.5 = 60 e^{- 0.1386 \times t}$
$e^{0.1386 \times t} = \frac{60}{7.5}$
$e^{0.1386 \times t} = 8$
Taking $\log$ both side, we get
$0.1386 \times t = 3 \ln 2$
$t = \frac{3 \times 0.693}{0.1386} = 15 s$

TS- EAMCET-10.09.2020

NUCLEAR PHYSICS

147581 In the Uranium radioactive series, the initial nucleus is $_{92}^{238} U$ and final nucleus is $_{82}^{206} Pb$ . When the Uranium nucleus decays to lead, the number of $α$ -particles emitted is ........ and the number of $β$ -particles emitted is

1 6,8

2 8,6

3 16,6

4 32,2

Explanation:

B When an a particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4
The number of alpha particles emitted is $\frac{238 - 206}{4} = \frac{32}{4} = 8$
When 8 alpha particles are emitted the atomic number decreases by $2 \times 8 = 16$ from 92 to $92 - 16 = 76$
When $β$ -particles is emitted the atomic number increases by 1 .
The number of $β$ -particles emitted $= 82 - 76 = 6$ .
Hence, the number of $α$ -particle emitted is 8 and the number of $β$ -particle emitted is 6 .

AP EAMCET (18.09.2020) Shift-II

NUCLEAR PHYSICS

147583 A radioactive source has a half-life of $6 h$ . A freshly prepared sample of the same exhibits radioactivity 32 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is

1

30 h

2

24 h

3

18 h

4

12 h

Explanation:

A Let's say the permissible level is N, so initially we have $32 N$ nuclei present.
We need to calculate the time taken to drop the number of nuclei from $32 N$ to $N$ .
So, $N = N_{0} e^{- λ t}$
$N_{o}$ is $32 N$ so after substituting this we get,
$e^{- λ t} = \frac{1}{32}$
$e^{λ t} = 32$
$λ t = \ln 32$
$t = \frac{\ln 32}{λ}$
We have, half life $(t_{1 / 2}) = 6 h$
$λ = \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 2}{6}$
Putting value of equation (ii) in equation (i), we get-
$t = \frac{\ln 32}{\ln 2} \times 6$
$t = \frac{5 \ln 2}{\ln 2} \times 6$
$t = 30 hours$

TS- EAMCET-09.09.2020

NUCLEAR PHYSICS

147584 The half-life of a radioactive isotope is $30 h$ . How long will it take to get reduced to $12.5 %$ of its initial amount?

1

120 h

2

90 h

3

60 h

4

50 h

Explanation:

B Given that,
Half life of radioactive sample $(T_{1 / 2}) = 30 h$
After decay, the remaining amount is $N$ -
So, $N = 12.5 %$ of $N_{0}$
$N = \frac{125}{1000} \times N_{0}$
$N = \frac{N_{0}}{8}$
As we know that,
$N = N_{0} {(\frac{1}{2})}^{n}$
From the equation (i) and equation (ii) we have-
$\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{n}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n}$
$n = 3$
Then,
$\frac{t}{T_{1 / 2}} = 3$
$\frac{t}{30} = 3$
$t = 90 h$

TS- EAMCET.14.09.2020

NUCLEAR PHYSICS

147585 Half-life of radioactive sample is $24 h$ . If a newly prepared radioactive sample shows 4 times the allowed and safe value of radio activity, the minimum time after which one can work safety with the source is

1

48 h

2

96 h

3

8 h

4

72 h

Explanation:

A Given, half life $(T_{1 / 2}) = 24 h$
According to question -
$\frac{N_{0}}{4 N_{o}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{t}{T_{1 / 2}} = 2$
$t = 2 \times T_{1 / 2}$
$t = 2 \times 24$
$t = 48 h$

TS- EAMCET-14.09.2020

NUCLEAR PHYSICS

147586 The half-life of a radioactive sample is $5 s$ . If the initial mass of the sample is $60 g$ , then the time required to reduce the sample to $7.5 g$ is

1

15 s

2

75 s

3

7.5 s

4

10 s

Explanation:

A Given,
Half life $(t_{1 / 2}) = 5 s$
Initial mass of sample $(N_{o}) = 60 g$
Reduce mass of sample $(N) = 7.5 g$
As we know,
$Half life (t_{1 / 2}) = \frac{\ln 2}{λ}$
$λ = \frac{0.693}{5} = 0.1386$
According to radioactivity decay law-
$N = N_{o} e^{- λ t}$
$7.5 = 60 e^{- 0.1386 \times t}$
$e^{0.1386 \times t} = \frac{60}{7.5}$
$e^{0.1386 \times t} = 8$
Taking $\log$ both side, we get
$0.1386 \times t = 3 \ln 2$
$t = \frac{3 \times 0.693}{0.1386} = 15 s$

TS- EAMCET-10.09.2020

NUCLEAR PHYSICS

147581 In the Uranium radioactive series, the initial nucleus is $_{92}^{238} U$ and final nucleus is $_{82}^{206} Pb$ . When the Uranium nucleus decays to lead, the number of $α$ -particles emitted is ........ and the number of $β$ -particles emitted is

1 6,8

2 8,6

3 16,6

4 32,2

Explanation:

B When an a particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4
The number of alpha particles emitted is $\frac{238 - 206}{4} = \frac{32}{4} = 8$
When 8 alpha particles are emitted the atomic number decreases by $2 \times 8 = 16$ from 92 to $92 - 16 = 76$
When $β$ -particles is emitted the atomic number increases by 1 .
The number of $β$ -particles emitted $= 82 - 76 = 6$ .
Hence, the number of $α$ -particle emitted is 8 and the number of $β$ -particle emitted is 6 .

AP EAMCET (18.09.2020) Shift-II

NUCLEAR PHYSICS

147583 A radioactive source has a half-life of $6 h$ . A freshly prepared sample of the same exhibits radioactivity 32 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is

1

30 h

2

24 h

3

18 h

4

12 h

Explanation:

A Let's say the permissible level is N, so initially we have $32 N$ nuclei present.
We need to calculate the time taken to drop the number of nuclei from $32 N$ to $N$ .
So, $N = N_{0} e^{- λ t}$
$N_{o}$ is $32 N$ so after substituting this we get,
$e^{- λ t} = \frac{1}{32}$
$e^{λ t} = 32$
$λ t = \ln 32$
$t = \frac{\ln 32}{λ}$
We have, half life $(t_{1 / 2}) = 6 h$
$λ = \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 2}{6}$
Putting value of equation (ii) in equation (i), we get-
$t = \frac{\ln 32}{\ln 2} \times 6$
$t = \frac{5 \ln 2}{\ln 2} \times 6$
$t = 30 hours$

TS- EAMCET-09.09.2020

NUCLEAR PHYSICS

147584 The half-life of a radioactive isotope is $30 h$ . How long will it take to get reduced to $12.5 %$ of its initial amount?

1

120 h

2

90 h

3

60 h

4

50 h

Explanation:

B Given that,
Half life of radioactive sample $(T_{1 / 2}) = 30 h$
After decay, the remaining amount is $N$ -
So, $N = 12.5 %$ of $N_{0}$
$N = \frac{125}{1000} \times N_{0}$
$N = \frac{N_{0}}{8}$
As we know that,
$N = N_{0} {(\frac{1}{2})}^{n}$
From the equation (i) and equation (ii) we have-
$\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{n}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n}$
$n = 3$
Then,
$\frac{t}{T_{1 / 2}} = 3$
$\frac{t}{30} = 3$
$t = 90 h$

TS- EAMCET.14.09.2020

NUCLEAR PHYSICS

147585 Half-life of radioactive sample is $24 h$ . If a newly prepared radioactive sample shows 4 times the allowed and safe value of radio activity, the minimum time after which one can work safety with the source is

1

48 h

2

96 h

3

8 h

4

72 h

Explanation:

A Given, half life $(T_{1 / 2}) = 24 h$
According to question -
$\frac{N_{0}}{4 N_{o}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{t}{T_{1 / 2}} = 2$
$t = 2 \times T_{1 / 2}$
$t = 2 \times 24$
$t = 48 h$

TS- EAMCET-14.09.2020

NUCLEAR PHYSICS

147586 The half-life of a radioactive sample is $5 s$ . If the initial mass of the sample is $60 g$ , then the time required to reduce the sample to $7.5 g$ is

1

15 s

2

75 s

3

7.5 s

4

10 s

Explanation:

A Given,
Half life $(t_{1 / 2}) = 5 s$
Initial mass of sample $(N_{o}) = 60 g$
Reduce mass of sample $(N) = 7.5 g$
As we know,
$Half life (t_{1 / 2}) = \frac{\ln 2}{λ}$
$λ = \frac{0.693}{5} = 0.1386$
According to radioactivity decay law-
$N = N_{o} e^{- λ t}$
$7.5 = 60 e^{- 0.1386 \times t}$
$e^{0.1386 \times t} = \frac{60}{7.5}$
$e^{0.1386 \times t} = 8$
Taking $\log$ both side, we get
$0.1386 \times t = 3 \ln 2$
$t = \frac{3 \times 0.693}{0.1386} = 15 s$

TS- EAMCET-10.09.2020

NUCLEAR PHYSICS

147581 In the Uranium radioactive series, the initial nucleus is $_{92}^{238} U$ and final nucleus is $_{82}^{206} Pb$ . When the Uranium nucleus decays to lead, the number of $α$ -particles emitted is ........ and the number of $β$ -particles emitted is

1 6,8

2 8,6

3 16,6

4 32,2

Explanation:

B When an a particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4
The number of alpha particles emitted is $\frac{238 - 206}{4} = \frac{32}{4} = 8$
When 8 alpha particles are emitted the atomic number decreases by $2 \times 8 = 16$ from 92 to $92 - 16 = 76$
When $β$ -particles is emitted the atomic number increases by 1 .
The number of $β$ -particles emitted $= 82 - 76 = 6$ .
Hence, the number of $α$ -particle emitted is 8 and the number of $β$ -particle emitted is 6 .

AP EAMCET (18.09.2020) Shift-II

NUCLEAR PHYSICS

147583 A radioactive source has a half-life of $6 h$ . A freshly prepared sample of the same exhibits radioactivity 32 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is

1

30 h

2

24 h

3

18 h

4

12 h

Explanation:

A Let's say the permissible level is N, so initially we have $32 N$ nuclei present.
We need to calculate the time taken to drop the number of nuclei from $32 N$ to $N$ .
So, $N = N_{0} e^{- λ t}$
$N_{o}$ is $32 N$ so after substituting this we get,
$e^{- λ t} = \frac{1}{32}$
$e^{λ t} = 32$
$λ t = \ln 32$
$t = \frac{\ln 32}{λ}$
We have, half life $(t_{1 / 2}) = 6 h$
$λ = \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 2}{6}$
Putting value of equation (ii) in equation (i), we get-
$t = \frac{\ln 32}{\ln 2} \times 6$
$t = \frac{5 \ln 2}{\ln 2} \times 6$
$t = 30 hours$

TS- EAMCET-09.09.2020

NUCLEAR PHYSICS

147584 The half-life of a radioactive isotope is $30 h$ . How long will it take to get reduced to $12.5 %$ of its initial amount?

1

120 h

2

90 h

3

60 h

4

50 h

Explanation:

B Given that,
Half life of radioactive sample $(T_{1 / 2}) = 30 h$
After decay, the remaining amount is $N$ -
So, $N = 12.5 %$ of $N_{0}$
$N = \frac{125}{1000} \times N_{0}$
$N = \frac{N_{0}}{8}$
As we know that,
$N = N_{0} {(\frac{1}{2})}^{n}$
From the equation (i) and equation (ii) we have-
$\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{n}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n}$
$n = 3$
Then,
$\frac{t}{T_{1 / 2}} = 3$
$\frac{t}{30} = 3$
$t = 90 h$

TS- EAMCET.14.09.2020

NUCLEAR PHYSICS

147585 Half-life of radioactive sample is $24 h$ . If a newly prepared radioactive sample shows 4 times the allowed and safe value of radio activity, the minimum time after which one can work safety with the source is

1

48 h

2

96 h

3

8 h

4

72 h

Explanation:

A Given, half life $(T_{1 / 2}) = 24 h$
According to question -
$\frac{N_{0}}{4 N_{o}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{t}{T_{1 / 2}} = 2$
$t = 2 \times T_{1 / 2}$
$t = 2 \times 24$
$t = 48 h$

TS- EAMCET-14.09.2020

NUCLEAR PHYSICS

147586 The half-life of a radioactive sample is $5 s$ . If the initial mass of the sample is $60 g$ , then the time required to reduce the sample to $7.5 g$ is

1

15 s

2

75 s

3

7.5 s

4

10 s

Explanation:

A Given,
Half life $(t_{1 / 2}) = 5 s$
Initial mass of sample $(N_{o}) = 60 g$
Reduce mass of sample $(N) = 7.5 g$
As we know,
$Half life (t_{1 / 2}) = \frac{\ln 2}{λ}$
$λ = \frac{0.693}{5} = 0.1386$
According to radioactivity decay law-
$N = N_{o} e^{- λ t}$
$7.5 = 60 e^{- 0.1386 \times t}$
$e^{0.1386 \times t} = \frac{60}{7.5}$
$e^{0.1386 \times t} = 8$
Taking $\log$ both side, we get
$0.1386 \times t = 3 \ln 2$
$t = \frac{3 \times 0.693}{0.1386} = 15 s$

TS- EAMCET-10.09.2020

NUCLEAR PHYSICS

147581 In the Uranium radioactive series, the initial nucleus is $_{92}^{238} U$ and final nucleus is $_{82}^{206} Pb$ . When the Uranium nucleus decays to lead, the number of $α$ -particles emitted is ........ and the number of $β$ -particles emitted is

1 6,8

2 8,6

3 16,6

4 32,2

Explanation:

B When an a particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4
The number of alpha particles emitted is $\frac{238 - 206}{4} = \frac{32}{4} = 8$
When 8 alpha particles are emitted the atomic number decreases by $2 \times 8 = 16$ from 92 to $92 - 16 = 76$
When $β$ -particles is emitted the atomic number increases by 1 .
The number of $β$ -particles emitted $= 82 - 76 = 6$ .
Hence, the number of $α$ -particle emitted is 8 and the number of $β$ -particle emitted is 6 .

AP EAMCET (18.09.2020) Shift-II

NUCLEAR PHYSICS

147583 A radioactive source has a half-life of $6 h$ . A freshly prepared sample of the same exhibits radioactivity 32 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is

1

30 h

2

24 h

3

18 h

4

12 h

Explanation:

A Let's say the permissible level is N, so initially we have $32 N$ nuclei present.
We need to calculate the time taken to drop the number of nuclei from $32 N$ to $N$ .
So, $N = N_{0} e^{- λ t}$
$N_{o}$ is $32 N$ so after substituting this we get,
$e^{- λ t} = \frac{1}{32}$
$e^{λ t} = 32$
$λ t = \ln 32$
$t = \frac{\ln 32}{λ}$
We have, half life $(t_{1 / 2}) = 6 h$
$λ = \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 2}{6}$
Putting value of equation (ii) in equation (i), we get-
$t = \frac{\ln 32}{\ln 2} \times 6$
$t = \frac{5 \ln 2}{\ln 2} \times 6$
$t = 30 hours$

TS- EAMCET-09.09.2020

NUCLEAR PHYSICS

147584 The half-life of a radioactive isotope is $30 h$ . How long will it take to get reduced to $12.5 %$ of its initial amount?

1

120 h

2

90 h

3

60 h

4

50 h

Explanation:

B Given that,
Half life of radioactive sample $(T_{1 / 2}) = 30 h$
After decay, the remaining amount is $N$ -
So, $N = 12.5 %$ of $N_{0}$
$N = \frac{125}{1000} \times N_{0}$
$N = \frac{N_{0}}{8}$
As we know that,
$N = N_{0} {(\frac{1}{2})}^{n}$
From the equation (i) and equation (ii) we have-
$\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{n}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n}$
$n = 3$
Then,
$\frac{t}{T_{1 / 2}} = 3$
$\frac{t}{30} = 3$
$t = 90 h$

TS- EAMCET.14.09.2020

NUCLEAR PHYSICS

147585 Half-life of radioactive sample is $24 h$ . If a newly prepared radioactive sample shows 4 times the allowed and safe value of radio activity, the minimum time after which one can work safety with the source is

1

48 h

2

96 h

3

8 h

4

72 h

Explanation:

A Given, half life $(T_{1 / 2}) = 24 h$
According to question -
$\frac{N_{0}}{4 N_{o}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{t}{T_{1 / 2}} = 2$
$t = 2 \times T_{1 / 2}$
$t = 2 \times 24$
$t = 48 h$

TS- EAMCET-14.09.2020

NUCLEAR PHYSICS

147586 The half-life of a radioactive sample is $5 s$ . If the initial mass of the sample is $60 g$ , then the time required to reduce the sample to $7.5 g$ is

1

15 s

2

75 s

3

7.5 s

4

10 s

Explanation:

A Given,
Half life $(t_{1 / 2}) = 5 s$
Initial mass of sample $(N_{o}) = 60 g$
Reduce mass of sample $(N) = 7.5 g$
As we know,
$Half life (t_{1 / 2}) = \frac{\ln 2}{λ}$
$λ = \frac{0.693}{5} = 0.1386$
According to radioactivity decay law-
$N = N_{o} e^{- λ t}$
$7.5 = 60 e^{- 0.1386 \times t}$
$e^{0.1386 \times t} = \frac{60}{7.5}$
$e^{0.1386 \times t} = 8$
Taking $\log$ both side, we get
$0.1386 \times t = 3 \ln 2$
$t = \frac{3 \times 0.693}{0.1386} = 15 s$

TS- EAMCET-10.09.2020