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NUCLEAR PHYSICS

147589 A radioactive nucleus can decay in two different processes with half life $0.7 hr$ and 0.3 hr.
The effective average life of the nucleus in minutes approximately is
(value of $\ln 2 = 0.7$ )

1 14

2 18

3 24

4 26

Explanation:

B Given,
${(T_{1 / 2})}_{1} = 0.7 hr$
${(T_{1 / 2})}_{2} = 0.3 hr$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{{(T_{1 / 2})}_{1}} + \frac{1}{{(T_{1 / 2})}_{2}}$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{0.7} + \frac{1}{0.3}$
${(T_{1 / 2})}_{e} = \frac{0.7 \times 0.3}{0.7 + 0.3} = \frac{0.21}{1}$
${(T_{1 / 2})}_{e} = 0.21 hr$
${(T_{1 / 2})}_{e} = 0.21 \times 60 \min$
$So, average life (T_{avg}) = \frac{1}{λ} = \frac{{(T_{1 / 2})}_{e}}{\ln 2}$
$= \frac{0.21 \times 60}{0.6932}$
$= \frac{0.21 \times 60}{0.7}$
$= 18 \min$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147590 An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and 20 yrs respectively. After time t, the alloy was found to consist of $(\frac{1}{e}) kg$ of $A$ and $1 kg$ of $B$ . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2 = 0.7$ )

1

(\frac{200}{7}) yrs

2

(\frac{10}{7}) yrs

3

7 yrs

4 70 yrs

Explanation:

A Given,
Initial amount of radioactive element $A =$ Initial amount of radioactive element $B$
${(N_{o})}_{A} = {(N_{o})}_{B}$
Half life of element $A, {(T_{1 / 2})}_{A} = 10 yr$
Half life of element $B, {(T_{1 / 2})}_{B} = 20 yr$
After time, $t$ , remaining amount of element $A$ is -
$N_{A} = \frac{1}{e} kg$
Remaining amount of element $B$ is -
$N_{B} = 1 kg$
For element A, after time t, amount of element left undecayed is given as -
$N_{A} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{{(T_{1 / 2})}_{A}}}$
$\frac{1}{e} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{10}}$
Similarly for element B,
$N_{B} = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
$1 = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
Dividing equation (ii) and (iii) and using equation (i), we get-
$\frac{1}{e} = \frac{{(\frac{1}{2})}^{\frac{t}{10}}}{{(\frac{1}{2})}^{\frac{t}{20}}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{10} - \frac{t}{20}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{20}}$
Taking $\log$ on the both sides, we get -
$\log \frac{1}{e} = \frac{t}{20} \log (\frac{1}{2})$
$t = \frac{20}{\log 2}$
$t = \frac{20}{0.7} [∵ \log 2 = 0.7]$
$So, t = \frac{200}{7} yr$

TS- EAMCET-03.05.2019

NUCLEAR PHYSICS

147591 The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 eV$ , travels a distance of $1 km$ ?
(mass of neutron $= 1.68 \times 10^{- 27} kg$ , and In $2 =$ 0.693)

1

60 \times 10^{- 5}

2

15 \times 10^{- 5}

3

25 \times 10^{- 5}

4

50 \times 10^{- 5}

Explanation:

C Given, half life $(T_{1 / 2}) = 693 \sec$
Kinetic energy $= 0.084 eV = 0.084 \times 1.6 \times 10^{- 19} J$
Distance $= 1 km = 1000 m$
The velocity (v) of neutron is given by -
$\frac{1}{2} {mv}^{2} = 0.084 eV$
$\frac{1}{2} \times 1.68 \times 10^{- 27} \times v^{2} = 0.084 eV$
$v^{2} = \frac{0.084 \times 2 \times 1.6 \times 10^{- 19}}{1.68 \times 10^{- 27}}$
$v^{2} = 0.16 \times 10^{8}$
$v^{2} = 16 \times 10^{6}$
$v = 4 \times 10^{3} m / s$
Time taken by neutron to travel a distance of $1 km$ ,
$Time (t) = \frac{Distance}{Velocity}$
$= \frac{1000}{4 \times 10^{3}} = 0.25 \sec$
And, half life $(T_{1 / 2}) = 693 s$
According to radioactive decay law,
$N = N_{o} e^{- λ t}$
$Then, \frac{N_{0}}{2} = N_{0} e^{- λ T_{1 / 2}}$
$\frac{1}{2} = e^{- λ T_{1 / 2}}$
$\log 2 = λ T_{1 / 2}$
$λ = \frac{0.693}{693}$
$λ = 1 \times 10^{- 3} s^{- 1}$
The fraction of neutron decayed in time $t$ is-
$\log (\frac{N}{N_{0}}) = λ t$
$\log \frac{N}{N_{0}} = 1 \times 10^{- 3} \times 0.25$
$= 25 \times 10^{- 5}$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147593 Two radioactive nuclei ' $x$ ' and ' $y$ ' initially contain equal number of atoms. Their half-life periods are 1 hour and 2 hours respectively. The ratio of their rate of disintegration after 2 hours from the start is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 3

Explanation:

B $x = N_{0} e^{- λ_{1} t}$
$y = N_{0} e^{- λ_{2} t^{t}}$
$\frac{x}{y} = e^{(λ_{2} - λ_{1}) t}$
Now, $λ_{1} = \frac{\ln 2}{1}$
$λ_{2} = \frac{\ln 2}{2}$
Putting value of $λ_{1}$ and $λ_{2}$ in equation (i), we get
$\frac{x}{y} = e^{(\frac{\ln 2}{2} - \frac{\ln 2}{1}) \times 2} [∵ t = 2 hours]$
$\frac{x}{y} = e^{- \ln 2}$
$\frac{x}{y} = \frac{1}{2} or 1 : 2$

AP EAMCET-23.04.2019

NUCLEAR PHYSICS

147589 A radioactive nucleus can decay in two different processes with half life $0.7 hr$ and 0.3 hr.
The effective average life of the nucleus in minutes approximately is
(value of $\ln 2 = 0.7$ )

1 14

2 18

3 24

4 26

Explanation:

B Given,
${(T_{1 / 2})}_{1} = 0.7 hr$
${(T_{1 / 2})}_{2} = 0.3 hr$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{{(T_{1 / 2})}_{1}} + \frac{1}{{(T_{1 / 2})}_{2}}$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{0.7} + \frac{1}{0.3}$
${(T_{1 / 2})}_{e} = \frac{0.7 \times 0.3}{0.7 + 0.3} = \frac{0.21}{1}$
${(T_{1 / 2})}_{e} = 0.21 hr$
${(T_{1 / 2})}_{e} = 0.21 \times 60 \min$
$So, average life (T_{avg}) = \frac{1}{λ} = \frac{{(T_{1 / 2})}_{e}}{\ln 2}$
$= \frac{0.21 \times 60}{0.6932}$
$= \frac{0.21 \times 60}{0.7}$
$= 18 \min$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147590 An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and 20 yrs respectively. After time t, the alloy was found to consist of $(\frac{1}{e}) kg$ of $A$ and $1 kg$ of $B$ . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2 = 0.7$ )

1

(\frac{200}{7}) yrs

2

(\frac{10}{7}) yrs

3

7 yrs

4 70 yrs

Explanation:

A Given,
Initial amount of radioactive element $A =$ Initial amount of radioactive element $B$
${(N_{o})}_{A} = {(N_{o})}_{B}$
Half life of element $A, {(T_{1 / 2})}_{A} = 10 yr$
Half life of element $B, {(T_{1 / 2})}_{B} = 20 yr$
After time, $t$ , remaining amount of element $A$ is -
$N_{A} = \frac{1}{e} kg$
Remaining amount of element $B$ is -
$N_{B} = 1 kg$
For element A, after time t, amount of element left undecayed is given as -
$N_{A} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{{(T_{1 / 2})}_{A}}}$
$\frac{1}{e} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{10}}$
Similarly for element B,
$N_{B} = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
$1 = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
Dividing equation (ii) and (iii) and using equation (i), we get-
$\frac{1}{e} = \frac{{(\frac{1}{2})}^{\frac{t}{10}}}{{(\frac{1}{2})}^{\frac{t}{20}}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{10} - \frac{t}{20}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{20}}$
Taking $\log$ on the both sides, we get -
$\log \frac{1}{e} = \frac{t}{20} \log (\frac{1}{2})$
$t = \frac{20}{\log 2}$
$t = \frac{20}{0.7} [∵ \log 2 = 0.7]$
$So, t = \frac{200}{7} yr$

TS- EAMCET-03.05.2019

NUCLEAR PHYSICS

147591 The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 eV$ , travels a distance of $1 km$ ?
(mass of neutron $= 1.68 \times 10^{- 27} kg$ , and In $2 =$ 0.693)

1

60 \times 10^{- 5}

2

15 \times 10^{- 5}

3

25 \times 10^{- 5}

4

50 \times 10^{- 5}

Explanation:

C Given, half life $(T_{1 / 2}) = 693 \sec$
Kinetic energy $= 0.084 eV = 0.084 \times 1.6 \times 10^{- 19} J$
Distance $= 1 km = 1000 m$
The velocity (v) of neutron is given by -
$\frac{1}{2} {mv}^{2} = 0.084 eV$
$\frac{1}{2} \times 1.68 \times 10^{- 27} \times v^{2} = 0.084 eV$
$v^{2} = \frac{0.084 \times 2 \times 1.6 \times 10^{- 19}}{1.68 \times 10^{- 27}}$
$v^{2} = 0.16 \times 10^{8}$
$v^{2} = 16 \times 10^{6}$
$v = 4 \times 10^{3} m / s$
Time taken by neutron to travel a distance of $1 km$ ,
$Time (t) = \frac{Distance}{Velocity}$
$= \frac{1000}{4 \times 10^{3}} = 0.25 \sec$
And, half life $(T_{1 / 2}) = 693 s$
According to radioactive decay law,
$N = N_{o} e^{- λ t}$
$Then, \frac{N_{0}}{2} = N_{0} e^{- λ T_{1 / 2}}$
$\frac{1}{2} = e^{- λ T_{1 / 2}}$
$\log 2 = λ T_{1 / 2}$
$λ = \frac{0.693}{693}$
$λ = 1 \times 10^{- 3} s^{- 1}$
The fraction of neutron decayed in time $t$ is-
$\log (\frac{N}{N_{0}}) = λ t$
$\log \frac{N}{N_{0}} = 1 \times 10^{- 3} \times 0.25$
$= 25 \times 10^{- 5}$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147593 Two radioactive nuclei ' $x$ ' and ' $y$ ' initially contain equal number of atoms. Their half-life periods are 1 hour and 2 hours respectively. The ratio of their rate of disintegration after 2 hours from the start is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 3

Explanation:

B $x = N_{0} e^{- λ_{1} t}$
$y = N_{0} e^{- λ_{2} t^{t}}$
$\frac{x}{y} = e^{(λ_{2} - λ_{1}) t}$
Now, $λ_{1} = \frac{\ln 2}{1}$
$λ_{2} = \frac{\ln 2}{2}$
Putting value of $λ_{1}$ and $λ_{2}$ in equation (i), we get
$\frac{x}{y} = e^{(\frac{\ln 2}{2} - \frac{\ln 2}{1}) \times 2} [∵ t = 2 hours]$
$\frac{x}{y} = e^{- \ln 2}$
$\frac{x}{y} = \frac{1}{2} or 1 : 2$

AP EAMCET-23.04.2019

NUCLEAR PHYSICS

147589 A radioactive nucleus can decay in two different processes with half life $0.7 hr$ and 0.3 hr.
The effective average life of the nucleus in minutes approximately is
(value of $\ln 2 = 0.7$ )

1 14

2 18

3 24

4 26

Explanation:

B Given,
${(T_{1 / 2})}_{1} = 0.7 hr$
${(T_{1 / 2})}_{2} = 0.3 hr$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{{(T_{1 / 2})}_{1}} + \frac{1}{{(T_{1 / 2})}_{2}}$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{0.7} + \frac{1}{0.3}$
${(T_{1 / 2})}_{e} = \frac{0.7 \times 0.3}{0.7 + 0.3} = \frac{0.21}{1}$
${(T_{1 / 2})}_{e} = 0.21 hr$
${(T_{1 / 2})}_{e} = 0.21 \times 60 \min$
$So, average life (T_{avg}) = \frac{1}{λ} = \frac{{(T_{1 / 2})}_{e}}{\ln 2}$
$= \frac{0.21 \times 60}{0.6932}$
$= \frac{0.21 \times 60}{0.7}$
$= 18 \min$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147590 An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and 20 yrs respectively. After time t, the alloy was found to consist of $(\frac{1}{e}) kg$ of $A$ and $1 kg$ of $B$ . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2 = 0.7$ )

1

(\frac{200}{7}) yrs

2

(\frac{10}{7}) yrs

3

7 yrs

4 70 yrs

Explanation:

A Given,
Initial amount of radioactive element $A =$ Initial amount of radioactive element $B$
${(N_{o})}_{A} = {(N_{o})}_{B}$
Half life of element $A, {(T_{1 / 2})}_{A} = 10 yr$
Half life of element $B, {(T_{1 / 2})}_{B} = 20 yr$
After time, $t$ , remaining amount of element $A$ is -
$N_{A} = \frac{1}{e} kg$
Remaining amount of element $B$ is -
$N_{B} = 1 kg$
For element A, after time t, amount of element left undecayed is given as -
$N_{A} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{{(T_{1 / 2})}_{A}}}$
$\frac{1}{e} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{10}}$
Similarly for element B,
$N_{B} = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
$1 = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
Dividing equation (ii) and (iii) and using equation (i), we get-
$\frac{1}{e} = \frac{{(\frac{1}{2})}^{\frac{t}{10}}}{{(\frac{1}{2})}^{\frac{t}{20}}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{10} - \frac{t}{20}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{20}}$
Taking $\log$ on the both sides, we get -
$\log \frac{1}{e} = \frac{t}{20} \log (\frac{1}{2})$
$t = \frac{20}{\log 2}$
$t = \frac{20}{0.7} [∵ \log 2 = 0.7]$
$So, t = \frac{200}{7} yr$

TS- EAMCET-03.05.2019

NUCLEAR PHYSICS

147591 The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 eV$ , travels a distance of $1 km$ ?
(mass of neutron $= 1.68 \times 10^{- 27} kg$ , and In $2 =$ 0.693)

1

60 \times 10^{- 5}

2

15 \times 10^{- 5}

3

25 \times 10^{- 5}

4

50 \times 10^{- 5}

Explanation:

C Given, half life $(T_{1 / 2}) = 693 \sec$
Kinetic energy $= 0.084 eV = 0.084 \times 1.6 \times 10^{- 19} J$
Distance $= 1 km = 1000 m$
The velocity (v) of neutron is given by -
$\frac{1}{2} {mv}^{2} = 0.084 eV$
$\frac{1}{2} \times 1.68 \times 10^{- 27} \times v^{2} = 0.084 eV$
$v^{2} = \frac{0.084 \times 2 \times 1.6 \times 10^{- 19}}{1.68 \times 10^{- 27}}$
$v^{2} = 0.16 \times 10^{8}$
$v^{2} = 16 \times 10^{6}$
$v = 4 \times 10^{3} m / s$
Time taken by neutron to travel a distance of $1 km$ ,
$Time (t) = \frac{Distance}{Velocity}$
$= \frac{1000}{4 \times 10^{3}} = 0.25 \sec$
And, half life $(T_{1 / 2}) = 693 s$
According to radioactive decay law,
$N = N_{o} e^{- λ t}$
$Then, \frac{N_{0}}{2} = N_{0} e^{- λ T_{1 / 2}}$
$\frac{1}{2} = e^{- λ T_{1 / 2}}$
$\log 2 = λ T_{1 / 2}$
$λ = \frac{0.693}{693}$
$λ = 1 \times 10^{- 3} s^{- 1}$
The fraction of neutron decayed in time $t$ is-
$\log (\frac{N}{N_{0}}) = λ t$
$\log \frac{N}{N_{0}} = 1 \times 10^{- 3} \times 0.25$
$= 25 \times 10^{- 5}$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147593 Two radioactive nuclei ' $x$ ' and ' $y$ ' initially contain equal number of atoms. Their half-life periods are 1 hour and 2 hours respectively. The ratio of their rate of disintegration after 2 hours from the start is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 3

Explanation:

B $x = N_{0} e^{- λ_{1} t}$
$y = N_{0} e^{- λ_{2} t^{t}}$
$\frac{x}{y} = e^{(λ_{2} - λ_{1}) t}$
Now, $λ_{1} = \frac{\ln 2}{1}$
$λ_{2} = \frac{\ln 2}{2}$
Putting value of $λ_{1}$ and $λ_{2}$ in equation (i), we get
$\frac{x}{y} = e^{(\frac{\ln 2}{2} - \frac{\ln 2}{1}) \times 2} [∵ t = 2 hours]$
$\frac{x}{y} = e^{- \ln 2}$
$\frac{x}{y} = \frac{1}{2} or 1 : 2$

AP EAMCET-23.04.2019

NUCLEAR PHYSICS

147589 A radioactive nucleus can decay in two different processes with half life $0.7 hr$ and 0.3 hr.
The effective average life of the nucleus in minutes approximately is
(value of $\ln 2 = 0.7$ )

1 14

2 18

3 24

4 26

Explanation:

B Given,
${(T_{1 / 2})}_{1} = 0.7 hr$
${(T_{1 / 2})}_{2} = 0.3 hr$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{{(T_{1 / 2})}_{1}} + \frac{1}{{(T_{1 / 2})}_{2}}$
$\frac{1}{{(T_{1 / 2})}_{e}} = \frac{1}{0.7} + \frac{1}{0.3}$
${(T_{1 / 2})}_{e} = \frac{0.7 \times 0.3}{0.7 + 0.3} = \frac{0.21}{1}$
${(T_{1 / 2})}_{e} = 0.21 hr$
${(T_{1 / 2})}_{e} = 0.21 \times 60 \min$
$So, average life (T_{avg}) = \frac{1}{λ} = \frac{{(T_{1 / 2})}_{e}}{\ln 2}$
$= \frac{0.21 \times 60}{0.6932}$
$= \frac{0.21 \times 60}{0.7}$
$= 18 \min$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147590 An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and 20 yrs respectively. After time t, the alloy was found to consist of $(\frac{1}{e}) kg$ of $A$ and $1 kg$ of $B$ . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2 = 0.7$ )

1

(\frac{200}{7}) yrs

2

(\frac{10}{7}) yrs

3

7 yrs

4 70 yrs

Explanation:

A Given,
Initial amount of radioactive element $A =$ Initial amount of radioactive element $B$
${(N_{o})}_{A} = {(N_{o})}_{B}$
Half life of element $A, {(T_{1 / 2})}_{A} = 10 yr$
Half life of element $B, {(T_{1 / 2})}_{B} = 20 yr$
After time, $t$ , remaining amount of element $A$ is -
$N_{A} = \frac{1}{e} kg$
Remaining amount of element $B$ is -
$N_{B} = 1 kg$
For element A, after time t, amount of element left undecayed is given as -
$N_{A} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{{(T_{1 / 2})}_{A}}}$
$\frac{1}{e} = {(N_{o})}_{A} {(\frac{1}{2})}^{\frac{t}{10}}$
Similarly for element B,
$N_{B} = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
$1 = {(N_{o})}_{B} {(\frac{1}{2})}^{\frac{t}{20}}$
Dividing equation (ii) and (iii) and using equation (i), we get-
$\frac{1}{e} = \frac{{(\frac{1}{2})}^{\frac{t}{10}}}{{(\frac{1}{2})}^{\frac{t}{20}}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{10} - \frac{t}{20}}$
$\frac{1}{e} = {(\frac{1}{2})}^{\frac{t}{20}}$
Taking $\log$ on the both sides, we get -
$\log \frac{1}{e} = \frac{t}{20} \log (\frac{1}{2})$
$t = \frac{20}{\log 2}$
$t = \frac{20}{0.7} [∵ \log 2 = 0.7]$
$So, t = \frac{200}{7} yr$

TS- EAMCET-03.05.2019

NUCLEAR PHYSICS

147591 The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 eV$ , travels a distance of $1 km$ ?
(mass of neutron $= 1.68 \times 10^{- 27} kg$ , and In $2 =$ 0.693)

1

60 \times 10^{- 5}

2

15 \times 10^{- 5}

3

25 \times 10^{- 5}

4

50 \times 10^{- 5}

Explanation:

C Given, half life $(T_{1 / 2}) = 693 \sec$
Kinetic energy $= 0.084 eV = 0.084 \times 1.6 \times 10^{- 19} J$
Distance $= 1 km = 1000 m$
The velocity (v) of neutron is given by -
$\frac{1}{2} {mv}^{2} = 0.084 eV$
$\frac{1}{2} \times 1.68 \times 10^{- 27} \times v^{2} = 0.084 eV$
$v^{2} = \frac{0.084 \times 2 \times 1.6 \times 10^{- 19}}{1.68 \times 10^{- 27}}$
$v^{2} = 0.16 \times 10^{8}$
$v^{2} = 16 \times 10^{6}$
$v = 4 \times 10^{3} m / s$
Time taken by neutron to travel a distance of $1 km$ ,
$Time (t) = \frac{Distance}{Velocity}$
$= \frac{1000}{4 \times 10^{3}} = 0.25 \sec$
And, half life $(T_{1 / 2}) = 693 s$
According to radioactive decay law,
$N = N_{o} e^{- λ t}$
$Then, \frac{N_{0}}{2} = N_{0} e^{- λ T_{1 / 2}}$
$\frac{1}{2} = e^{- λ T_{1 / 2}}$
$\log 2 = λ T_{1 / 2}$
$λ = \frac{0.693}{693}$
$λ = 1 \times 10^{- 3} s^{- 1}$
The fraction of neutron decayed in time $t$ is-
$\log (\frac{N}{N_{0}}) = λ t$
$\log \frac{N}{N_{0}} = 1 \times 10^{- 3} \times 0.25$
$= 25 \times 10^{- 5}$

TS- EAMCET-04.05.2019

NUCLEAR PHYSICS

147593 Two radioactive nuclei ' $x$ ' and ' $y$ ' initially contain equal number of atoms. Their half-life periods are 1 hour and 2 hours respectively. The ratio of their rate of disintegration after 2 hours from the start is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 3

Explanation:

B $x = N_{0} e^{- λ_{1} t}$
$y = N_{0} e^{- λ_{2} t^{t}}$
$\frac{x}{y} = e^{(λ_{2} - λ_{1}) t}$
Now, $λ_{1} = \frac{\ln 2}{1}$
$λ_{2} = \frac{\ln 2}{2}$
Putting value of $λ_{1}$ and $λ_{2}$ in equation (i), we get
$\frac{x}{y} = e^{(\frac{\ln 2}{2} - \frac{\ln 2}{1}) \times 2} [∵ t = 2 hours]$
$\frac{x}{y} = e^{- \ln 2}$
$\frac{x}{y} = \frac{1}{2} or 1 : 2$

AP EAMCET-23.04.2019