QBManager

PHXII13:NUCLEI

363983 If half-life of an element is $69.3 h$ , then how much of its percent will decay in $10 t h$ to $11 t h h$ ? (Initial activity $= 50 μ C i)$

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

Given, half-life of an element, $T_{1 / 2} = 69.3 h$
Let initial amount (at $t = 0$ ) of radioactive element is $N_{0}$
Active nuclei at $t = 10 h$ ,
$N_{1} = N_{0} e^{- 10 λ}$
Active nuclei at $t = 11 h, N_{2} = N_{0} e^{- 11 λ}$
$%$ decay in 10 th to 11 th hours
$= \frac{N_{1} - N_{2}}{N_{1}} \times 100$
$= \frac{N_{0} e^{- 10 λ} - N_{0} e^{- 11 λ}}{N_{0} e^{- 10 λ}} \times 100$
$= (1 - e^{- λ}) \times 100$
$= [1 - e^{\frac{- 0.693}{69.3}}] \times 100 [∵$ where, $λ = \frac{0.693}{T_{1 / 2}}]$
$\begin{aligned} = [1 - e^{- 0.01}] \times 100 \\ = [1 - 0.99] \times 100 \\ = 0.01 \times 100 = 1 % \end{aligned}$

AIIMS - 2019

PHXII13:NUCLEI

363984 Substance $A$ has atomic mass number 16 and half life of 1 day. Another substance $B$ has atomic mass number 32 and half life of $1 / 2$ day. If both $A$ and $B$ simultaneously start undergo radioactivity at the same time with initial mass $320 g$ each, how many total atoms of $A$ and $B$ combined would be left after 2 days.

1

3.38 \times 10^{24}

2

6.76 \times 10^{24}

3

1.69 \times 10^{24}

4

6.76 \times 10^{23}

Explanation:

Given: Mass number of $A = 16, T_{1 / 2} = 1$ day
Mass number of $B = 32, T_{1 / 2} = 1 / 2$ day
Initial mass $= 320 g, t = 2$ days
Number of initial moles of $A$ and $B$ are given by
$N_{0 (A)} = \frac{320}{16} = 20$ moles.
$N_{0 (B)} = \frac{320}{32} = 10$ moles.
As $n = \frac{t}{T_{1 / 2}}$
$N_{(A)} = N_{0 (A)} {(\frac{1}{2})}^{n} = 20 {(\frac{1}{2})}^{2 / 1} = 5$
$N_{(B)} = N_{0 (B)} {(\frac{1}{2})}^{n} = 10 {(\frac{1}{2})}^{2 / 1 / 2} = 10 {(\frac{1}{2})}^{4}$ $= 0.625$
$N = N_{A} + N_{B} = 5 + 0.625 = 5.625$ moles
No. of atoms left $= 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24}$

JEE - 2023

PHXII13:NUCLEI

363985 A radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample of $S_{2}$ activity $A_{2}$ . If $A_{2} = 2 A_{1}$ , then the ratio of half life of $S_{1}$ to the life of $S_{2}$ is

1

4

2

2

3

0.25

4

0.75

Explanation:

Activity , $A = λ N = \frac{0.693}{T} N$
Where $T$ is the half life of a radioactive sample, $N$ is the number of nuclei and $λ$ is the decay constant.
$∴ \frac{A_{1}}{A_{2}} = \frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}} \times \frac{N_{1}}{N_{2}} = \frac{2 A_{1}}{A_{1}} \times \frac{2 N_{2}}{N_{2}} = \frac{4}{1}$

KCET - 2010

PHXII13:NUCLEI

363986 The fraction $f$ of radioactive material that has decayed in time $t$ , varies with time $t$ . The correct variation is given by the curve
supporting img

1

A

2

B

3

C

4

D

Explanation:

Number of atoms undecayed $N = N_{o} e^{- λ t}$ number of atoms decayed
$= N_{o} - N = N_{o} (1 - e^{- λ t})$
$\Rightarrow$ Fraction of atoms that got decayed in time $t$ is
$f = \frac{N_{o} - N}{N_{o}} = 1 - e^{- λ t}$
i.e., fraction will rise up to 1, following exponential path as shown for curve ( $B$ ).

PHXII13:NUCLEI

363983 If half-life of an element is $69.3 h$ , then how much of its percent will decay in $10 t h$ to $11 t h h$ ? (Initial activity $= 50 μ C i)$

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

Given, half-life of an element, $T_{1 / 2} = 69.3 h$
Let initial amount (at $t = 0$ ) of radioactive element is $N_{0}$
Active nuclei at $t = 10 h$ ,
$N_{1} = N_{0} e^{- 10 λ}$
Active nuclei at $t = 11 h, N_{2} = N_{0} e^{- 11 λ}$
$%$ decay in 10 th to 11 th hours
$= \frac{N_{1} - N_{2}}{N_{1}} \times 100$
$= \frac{N_{0} e^{- 10 λ} - N_{0} e^{- 11 λ}}{N_{0} e^{- 10 λ}} \times 100$
$= (1 - e^{- λ}) \times 100$
$= [1 - e^{\frac{- 0.693}{69.3}}] \times 100 [∵$ where, $λ = \frac{0.693}{T_{1 / 2}}]$
$\begin{aligned} = [1 - e^{- 0.01}] \times 100 \\ = [1 - 0.99] \times 100 \\ = 0.01 \times 100 = 1 % \end{aligned}$

AIIMS - 2019

PHXII13:NUCLEI

363984 Substance $A$ has atomic mass number 16 and half life of 1 day. Another substance $B$ has atomic mass number 32 and half life of $1 / 2$ day. If both $A$ and $B$ simultaneously start undergo radioactivity at the same time with initial mass $320 g$ each, how many total atoms of $A$ and $B$ combined would be left after 2 days.

1

3.38 \times 10^{24}

2

6.76 \times 10^{24}

3

1.69 \times 10^{24}

4

6.76 \times 10^{23}

Explanation:

Given: Mass number of $A = 16, T_{1 / 2} = 1$ day
Mass number of $B = 32, T_{1 / 2} = 1 / 2$ day
Initial mass $= 320 g, t = 2$ days
Number of initial moles of $A$ and $B$ are given by
$N_{0 (A)} = \frac{320}{16} = 20$ moles.
$N_{0 (B)} = \frac{320}{32} = 10$ moles.
As $n = \frac{t}{T_{1 / 2}}$
$N_{(A)} = N_{0 (A)} {(\frac{1}{2})}^{n} = 20 {(\frac{1}{2})}^{2 / 1} = 5$
$N_{(B)} = N_{0 (B)} {(\frac{1}{2})}^{n} = 10 {(\frac{1}{2})}^{2 / 1 / 2} = 10 {(\frac{1}{2})}^{4}$ $= 0.625$
$N = N_{A} + N_{B} = 5 + 0.625 = 5.625$ moles
No. of atoms left $= 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24}$

JEE - 2023

PHXII13:NUCLEI

363985 A radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample of $S_{2}$ activity $A_{2}$ . If $A_{2} = 2 A_{1}$ , then the ratio of half life of $S_{1}$ to the life of $S_{2}$ is

1

4

2

2

3

0.25

4

0.75

Explanation:

Activity , $A = λ N = \frac{0.693}{T} N$
Where $T$ is the half life of a radioactive sample, $N$ is the number of nuclei and $λ$ is the decay constant.
$∴ \frac{A_{1}}{A_{2}} = \frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}} \times \frac{N_{1}}{N_{2}} = \frac{2 A_{1}}{A_{1}} \times \frac{2 N_{2}}{N_{2}} = \frac{4}{1}$

KCET - 2010

PHXII13:NUCLEI

363986 The fraction $f$ of radioactive material that has decayed in time $t$ , varies with time $t$ . The correct variation is given by the curve
supporting img

1

A

2

B

3

C

4

D

Explanation:

Number of atoms undecayed $N = N_{o} e^{- λ t}$ number of atoms decayed
$= N_{o} - N = N_{o} (1 - e^{- λ t})$
$\Rightarrow$ Fraction of atoms that got decayed in time $t$ is
$f = \frac{N_{o} - N}{N_{o}} = 1 - e^{- λ t}$
i.e., fraction will rise up to 1, following exponential path as shown for curve ( $B$ ).

PHXII13:NUCLEI

363983 If half-life of an element is $69.3 h$ , then how much of its percent will decay in $10 t h$ to $11 t h h$ ? (Initial activity $= 50 μ C i)$

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

Given, half-life of an element, $T_{1 / 2} = 69.3 h$
Let initial amount (at $t = 0$ ) of radioactive element is $N_{0}$
Active nuclei at $t = 10 h$ ,
$N_{1} = N_{0} e^{- 10 λ}$
Active nuclei at $t = 11 h, N_{2} = N_{0} e^{- 11 λ}$
$%$ decay in 10 th to 11 th hours
$= \frac{N_{1} - N_{2}}{N_{1}} \times 100$
$= \frac{N_{0} e^{- 10 λ} - N_{0} e^{- 11 λ}}{N_{0} e^{- 10 λ}} \times 100$
$= (1 - e^{- λ}) \times 100$
$= [1 - e^{\frac{- 0.693}{69.3}}] \times 100 [∵$ where, $λ = \frac{0.693}{T_{1 / 2}}]$
$\begin{aligned} = [1 - e^{- 0.01}] \times 100 \\ = [1 - 0.99] \times 100 \\ = 0.01 \times 100 = 1 % \end{aligned}$

AIIMS - 2019

PHXII13:NUCLEI

363984 Substance $A$ has atomic mass number 16 and half life of 1 day. Another substance $B$ has atomic mass number 32 and half life of $1 / 2$ day. If both $A$ and $B$ simultaneously start undergo radioactivity at the same time with initial mass $320 g$ each, how many total atoms of $A$ and $B$ combined would be left after 2 days.

1

3.38 \times 10^{24}

2

6.76 \times 10^{24}

3

1.69 \times 10^{24}

4

6.76 \times 10^{23}

Explanation:

Given: Mass number of $A = 16, T_{1 / 2} = 1$ day
Mass number of $B = 32, T_{1 / 2} = 1 / 2$ day
Initial mass $= 320 g, t = 2$ days
Number of initial moles of $A$ and $B$ are given by
$N_{0 (A)} = \frac{320}{16} = 20$ moles.
$N_{0 (B)} = \frac{320}{32} = 10$ moles.
As $n = \frac{t}{T_{1 / 2}}$
$N_{(A)} = N_{0 (A)} {(\frac{1}{2})}^{n} = 20 {(\frac{1}{2})}^{2 / 1} = 5$
$N_{(B)} = N_{0 (B)} {(\frac{1}{2})}^{n} = 10 {(\frac{1}{2})}^{2 / 1 / 2} = 10 {(\frac{1}{2})}^{4}$ $= 0.625$
$N = N_{A} + N_{B} = 5 + 0.625 = 5.625$ moles
No. of atoms left $= 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24}$

JEE - 2023

PHXII13:NUCLEI

363985 A radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample of $S_{2}$ activity $A_{2}$ . If $A_{2} = 2 A_{1}$ , then the ratio of half life of $S_{1}$ to the life of $S_{2}$ is

1

4

2

2

3

0.25

4

0.75

Explanation:

Activity , $A = λ N = \frac{0.693}{T} N$
Where $T$ is the half life of a radioactive sample, $N$ is the number of nuclei and $λ$ is the decay constant.
$∴ \frac{A_{1}}{A_{2}} = \frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}} \times \frac{N_{1}}{N_{2}} = \frac{2 A_{1}}{A_{1}} \times \frac{2 N_{2}}{N_{2}} = \frac{4}{1}$

KCET - 2010

PHXII13:NUCLEI

363986 The fraction $f$ of radioactive material that has decayed in time $t$ , varies with time $t$ . The correct variation is given by the curve
supporting img

1

A

2

B

3

C

4

D

Explanation:

Number of atoms undecayed $N = N_{o} e^{- λ t}$ number of atoms decayed
$= N_{o} - N = N_{o} (1 - e^{- λ t})$
$\Rightarrow$ Fraction of atoms that got decayed in time $t$ is
$f = \frac{N_{o} - N}{N_{o}} = 1 - e^{- λ t}$
i.e., fraction will rise up to 1, following exponential path as shown for curve ( $B$ ).

PHXII13:NUCLEI

363983 If half-life of an element is $69.3 h$ , then how much of its percent will decay in $10 t h$ to $11 t h h$ ? (Initial activity $= 50 μ C i)$

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

Given, half-life of an element, $T_{1 / 2} = 69.3 h$
Let initial amount (at $t = 0$ ) of radioactive element is $N_{0}$
Active nuclei at $t = 10 h$ ,
$N_{1} = N_{0} e^{- 10 λ}$
Active nuclei at $t = 11 h, N_{2} = N_{0} e^{- 11 λ}$
$%$ decay in 10 th to 11 th hours
$= \frac{N_{1} - N_{2}}{N_{1}} \times 100$
$= \frac{N_{0} e^{- 10 λ} - N_{0} e^{- 11 λ}}{N_{0} e^{- 10 λ}} \times 100$
$= (1 - e^{- λ}) \times 100$
$= [1 - e^{\frac{- 0.693}{69.3}}] \times 100 [∵$ where, $λ = \frac{0.693}{T_{1 / 2}}]$
$\begin{aligned} = [1 - e^{- 0.01}] \times 100 \\ = [1 - 0.99] \times 100 \\ = 0.01 \times 100 = 1 % \end{aligned}$

AIIMS - 2019

PHXII13:NUCLEI

363984 Substance $A$ has atomic mass number 16 and half life of 1 day. Another substance $B$ has atomic mass number 32 and half life of $1 / 2$ day. If both $A$ and $B$ simultaneously start undergo radioactivity at the same time with initial mass $320 g$ each, how many total atoms of $A$ and $B$ combined would be left after 2 days.

1

3.38 \times 10^{24}

2

6.76 \times 10^{24}

3

1.69 \times 10^{24}

4

6.76 \times 10^{23}

Explanation:

Given: Mass number of $A = 16, T_{1 / 2} = 1$ day
Mass number of $B = 32, T_{1 / 2} = 1 / 2$ day
Initial mass $= 320 g, t = 2$ days
Number of initial moles of $A$ and $B$ are given by
$N_{0 (A)} = \frac{320}{16} = 20$ moles.
$N_{0 (B)} = \frac{320}{32} = 10$ moles.
As $n = \frac{t}{T_{1 / 2}}$
$N_{(A)} = N_{0 (A)} {(\frac{1}{2})}^{n} = 20 {(\frac{1}{2})}^{2 / 1} = 5$
$N_{(B)} = N_{0 (B)} {(\frac{1}{2})}^{n} = 10 {(\frac{1}{2})}^{2 / 1 / 2} = 10 {(\frac{1}{2})}^{4}$ $= 0.625$
$N = N_{A} + N_{B} = 5 + 0.625 = 5.625$ moles
No. of atoms left $= 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24}$

JEE - 2023

PHXII13:NUCLEI

363985 A radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample of $S_{2}$ activity $A_{2}$ . If $A_{2} = 2 A_{1}$ , then the ratio of half life of $S_{1}$ to the life of $S_{2}$ is

1

4

2

2

3

0.25

4

0.75

Explanation:

Activity , $A = λ N = \frac{0.693}{T} N$
Where $T$ is the half life of a radioactive sample, $N$ is the number of nuclei and $λ$ is the decay constant.
$∴ \frac{A_{1}}{A_{2}} = \frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}} \times \frac{N_{1}}{N_{2}} = \frac{2 A_{1}}{A_{1}} \times \frac{2 N_{2}}{N_{2}} = \frac{4}{1}$

KCET - 2010

PHXII13:NUCLEI

363986 The fraction $f$ of radioactive material that has decayed in time $t$ , varies with time $t$ . The correct variation is given by the curve
supporting img

1

A

2

B

3

C

4

D

Explanation:

Number of atoms undecayed $N = N_{o} e^{- λ t}$ number of atoms decayed
$= N_{o} - N = N_{o} (1 - e^{- λ t})$
$\Rightarrow$ Fraction of atoms that got decayed in time $t$ is
$f = \frac{N_{o} - N}{N_{o}} = 1 - e^{- λ t}$
i.e., fraction will rise up to 1, following exponential path as shown for curve ( $B$ ).

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