QBManager

NUCLEAR PHYSICS

147820 A radioactive substance of half life 138.6 days is placed in a box. After $n$ days only $20 %$ of the substance is present then the value of $n$ is $[\ln (5) = 1.6]$

1 693

2 138.6

3 277.2

4 322

Explanation:

D Given that,
$t_{1 / 2} = 138.6 day$
$λ = \frac{0.693}{138.6} = 5 \times 10^{- 3} \sec$
$t = n day$
According to question -
$N = 20 % N_{0}$
$N = \frac{20}{100} \times N_{0}$
$N = \frac{N_{0}}{5}$
We know that, $N = N_{0} e^{- λ t}$
$\frac{N_{0}}{5} = N_{0} e^{- λ n}$
$\frac{1}{5} = \frac{1}{e^{λ n}}$
$e^{λ n} = 5$
$λ n = \ln 5$
$n = \frac{1.61}{5 \times 10^{- 3}} (∵ \ln 5 = 1.61)$
$n = 322 day$

AP EAMCET (22.04.2019) Shift-I

NUCLEAR PHYSICS

147821 An atomic power nuclear reactor can deliver $300 MW$ . The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $MeV$ . The number of uranium atoms fashioned per hour will be

1

30 \times 10^{25}

2

4 \times 10^{22}

3

10 \times 10^{20}

4

5 \times 10^{15}

Explanation:

B Given that,
Power $(P) = 300 MW$
Energy $(E) = 170 MeV$
We know that,
$P = n \frac{E}{T}$
$300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$300 \times 10^{6} = \frac{n \times 170 \times 1.6 \times 10^{- 13}}{t}$
So, the number of atom per second $= \frac{n}{t} = 1.102 \times 10^{19}$ Hence, the number of atom per hour $= 1.102 \times 10^{19} \times$ $3600 = 3.97 \times 10^{22}$
The number of atom per hours $\approx 4 \times 10^{22}$ .

JIPMER-2018

NUCLEAR PHYSICS

147823 If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day?
(Given, $\sqrt[3]{0.25} \approx 0.63$ )

1 0.63

2 0.5

3 0.37

4 0.13

Explanation:

D Given that,
Half-life of radioactive nuclei $(T_{1 / 2}) = 3$ days
Number of active nuclei left alter $t$ ,
$N = N_{0} {(\frac{1}{2})}^{n} (∵ n = \frac{t}{T_{1 / 2}})$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
After $t = 2$ days then-
$N_{1} = N_{0} {(\frac{1}{2})}^{\frac{2}{3}} = \frac{N_{0}}{2^{2 / 3}} = \frac{N_{0}}{4^{1 / 3}} = 0.63 N_{0}$
$N_{1} = 0.63 N_{0}$
After $t = 3$ days
$N_{2} = N_{0} {(\frac{1}{2})}^{\frac{3}{3}} = \frac{N_{0}}{2} = 0.5 N_{0}$
Number of nuclei that will decay on the $3^{rd}$ day,
$N_{3} = N_{1} - N_{2}$
$N_{3} = 0.63 N_{o} - 0.5 N_{o}$
$N_{3} = 0.13 N_{o}$

WB JEE 2018

NUCLEAR PHYSICS

147824 The half-life of tritium is $12.5$ years. What mass of tritium of initial mass $64 mg$ will remain undecayed after 50 years ?

1

32 mg

2

8 mg

3

16 mg

4

4 mg

Explanation:

D Given that,
Half-life of tritium $(T_{1 / 2}) = 12.5$ years
Initial mass $(N_{0}) = 64 mg$ .
Total time $(t) = 50$ year From the radioactive equation-
$N = N_{0} {(\frac{1}{2})}^{n}$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$N = 64 {(\frac{1}{2})}^{\frac{50}{12.5}} \Rightarrow N = 64 \times {(\frac{1}{2})}^{4}$
$N = 64 \times \frac{1}{16}$
$N = 4 mg$

Karnataka CET-2018

NUCLEAR PHYSICS

147825 The luminous dials watches are usually made by mixing a zinc sulphide phosphor with an aparticle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 a - particles per second for this purpose is

1

2.77 mg

2

2.77 g

3

2.77 \times 10^{- 23} g

4

2.77 \times 10^{- 13} kg

.

Explanation:

D Given that,
The rate of radioactive decay $= 10$
$\frac{dN}{dt} = 10$
For the first order decay -
$\frac{dN}{dt} = λ N$
$10 = λ N$
$N = \frac{10}{λ} (∵ λ = \frac{\ln 2}{t_{1 / 2}})$
$N = \frac{10}{\frac{\ln 2}{t_{1 / 2}}}$
$N = \frac{10 \times t_{1 / 2}}{\ln 2}$
$N = \frac{10 \times [1620 \times (365 \times 24 \times 3600)]}{N = 7.3 \times 10^{11}}$
$\ln 2$
Atom is present in,
$= \frac{N}{N_{A}} = \frac{7.3 \times 10^{11}}{6.023 \times 10^{23}} = 1.223 \times 10^{- 12} moles$
According to question -
Atomic mass of radium $= 226$ mole
1 Mole of radium as a mass of $226 g$
Mass of radium $= 1.223 \times 10^{- 12} \times 226 = 2.765 \times 10^{- 10} g$
$= 2.765 \times 10^{- 13} kg$

AIIMS-26.05.2018(M)

NUCLEAR PHYSICS

147820 A radioactive substance of half life 138.6 days is placed in a box. After $n$ days only $20 %$ of the substance is present then the value of $n$ is $[\ln (5) = 1.6]$

1 693

2 138.6

3 277.2

4 322

Explanation:

D Given that,
$t_{1 / 2} = 138.6 day$
$λ = \frac{0.693}{138.6} = 5 \times 10^{- 3} \sec$
$t = n day$
According to question -
$N = 20 % N_{0}$
$N = \frac{20}{100} \times N_{0}$
$N = \frac{N_{0}}{5}$
We know that, $N = N_{0} e^{- λ t}$
$\frac{N_{0}}{5} = N_{0} e^{- λ n}$
$\frac{1}{5} = \frac{1}{e^{λ n}}$
$e^{λ n} = 5$
$λ n = \ln 5$
$n = \frac{1.61}{5 \times 10^{- 3}} (∵ \ln 5 = 1.61)$
$n = 322 day$

AP EAMCET (22.04.2019) Shift-I

NUCLEAR PHYSICS

147821 An atomic power nuclear reactor can deliver $300 MW$ . The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $MeV$ . The number of uranium atoms fashioned per hour will be

1

30 \times 10^{25}

2

4 \times 10^{22}

3

10 \times 10^{20}

4

5 \times 10^{15}

Explanation:

B Given that,
Power $(P) = 300 MW$
Energy $(E) = 170 MeV$
We know that,
$P = n \frac{E}{T}$
$300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$300 \times 10^{6} = \frac{n \times 170 \times 1.6 \times 10^{- 13}}{t}$
So, the number of atom per second $= \frac{n}{t} = 1.102 \times 10^{19}$ Hence, the number of atom per hour $= 1.102 \times 10^{19} \times$ $3600 = 3.97 \times 10^{22}$
The number of atom per hours $\approx 4 \times 10^{22}$ .

JIPMER-2018

NUCLEAR PHYSICS

147823 If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day?
(Given, $\sqrt[3]{0.25} \approx 0.63$ )

1 0.63

2 0.5

3 0.37

4 0.13

Explanation:

D Given that,
Half-life of radioactive nuclei $(T_{1 / 2}) = 3$ days
Number of active nuclei left alter $t$ ,
$N = N_{0} {(\frac{1}{2})}^{n} (∵ n = \frac{t}{T_{1 / 2}})$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
After $t = 2$ days then-
$N_{1} = N_{0} {(\frac{1}{2})}^{\frac{2}{3}} = \frac{N_{0}}{2^{2 / 3}} = \frac{N_{0}}{4^{1 / 3}} = 0.63 N_{0}$
$N_{1} = 0.63 N_{0}$
After $t = 3$ days
$N_{2} = N_{0} {(\frac{1}{2})}^{\frac{3}{3}} = \frac{N_{0}}{2} = 0.5 N_{0}$
Number of nuclei that will decay on the $3^{rd}$ day,
$N_{3} = N_{1} - N_{2}$
$N_{3} = 0.63 N_{o} - 0.5 N_{o}$
$N_{3} = 0.13 N_{o}$

WB JEE 2018

NUCLEAR PHYSICS

147824 The half-life of tritium is $12.5$ years. What mass of tritium of initial mass $64 mg$ will remain undecayed after 50 years ?

1

32 mg

2

8 mg

3

16 mg

4

4 mg

Explanation:

D Given that,
Half-life of tritium $(T_{1 / 2}) = 12.5$ years
Initial mass $(N_{0}) = 64 mg$ .
Total time $(t) = 50$ year From the radioactive equation-
$N = N_{0} {(\frac{1}{2})}^{n}$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$N = 64 {(\frac{1}{2})}^{\frac{50}{12.5}} \Rightarrow N = 64 \times {(\frac{1}{2})}^{4}$
$N = 64 \times \frac{1}{16}$
$N = 4 mg$

Karnataka CET-2018

NUCLEAR PHYSICS

147825 The luminous dials watches are usually made by mixing a zinc sulphide phosphor with an aparticle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 a - particles per second for this purpose is

1

2.77 mg

2

2.77 g

3

2.77 \times 10^{- 23} g

4

2.77 \times 10^{- 13} kg

.

Explanation:

D Given that,
The rate of radioactive decay $= 10$
$\frac{dN}{dt} = 10$
For the first order decay -
$\frac{dN}{dt} = λ N$
$10 = λ N$
$N = \frac{10}{λ} (∵ λ = \frac{\ln 2}{t_{1 / 2}})$
$N = \frac{10}{\frac{\ln 2}{t_{1 / 2}}}$
$N = \frac{10 \times t_{1 / 2}}{\ln 2}$
$N = \frac{10 \times [1620 \times (365 \times 24 \times 3600)]}{N = 7.3 \times 10^{11}}$
$\ln 2$
Atom is present in,
$= \frac{N}{N_{A}} = \frac{7.3 \times 10^{11}}{6.023 \times 10^{23}} = 1.223 \times 10^{- 12} moles$
According to question -
Atomic mass of radium $= 226$ mole
1 Mole of radium as a mass of $226 g$
Mass of radium $= 1.223 \times 10^{- 12} \times 226 = 2.765 \times 10^{- 10} g$
$= 2.765 \times 10^{- 13} kg$

AIIMS-26.05.2018(M)

NUCLEAR PHYSICS

147820 A radioactive substance of half life 138.6 days is placed in a box. After $n$ days only $20 %$ of the substance is present then the value of $n$ is $[\ln (5) = 1.6]$

1 693

2 138.6

3 277.2

4 322

Explanation:

D Given that,
$t_{1 / 2} = 138.6 day$
$λ = \frac{0.693}{138.6} = 5 \times 10^{- 3} \sec$
$t = n day$
According to question -
$N = 20 % N_{0}$
$N = \frac{20}{100} \times N_{0}$
$N = \frac{N_{0}}{5}$
We know that, $N = N_{0} e^{- λ t}$
$\frac{N_{0}}{5} = N_{0} e^{- λ n}$
$\frac{1}{5} = \frac{1}{e^{λ n}}$
$e^{λ n} = 5$
$λ n = \ln 5$
$n = \frac{1.61}{5 \times 10^{- 3}} (∵ \ln 5 = 1.61)$
$n = 322 day$

AP EAMCET (22.04.2019) Shift-I

NUCLEAR PHYSICS

147821 An atomic power nuclear reactor can deliver $300 MW$ . The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $MeV$ . The number of uranium atoms fashioned per hour will be

1

30 \times 10^{25}

2

4 \times 10^{22}

3

10 \times 10^{20}

4

5 \times 10^{15}

Explanation:

B Given that,
Power $(P) = 300 MW$
Energy $(E) = 170 MeV$
We know that,
$P = n \frac{E}{T}$
$300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$300 \times 10^{6} = \frac{n \times 170 \times 1.6 \times 10^{- 13}}{t}$
So, the number of atom per second $= \frac{n}{t} = 1.102 \times 10^{19}$ Hence, the number of atom per hour $= 1.102 \times 10^{19} \times$ $3600 = 3.97 \times 10^{22}$
The number of atom per hours $\approx 4 \times 10^{22}$ .

JIPMER-2018

NUCLEAR PHYSICS

147823 If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day?
(Given, $\sqrt[3]{0.25} \approx 0.63$ )

1 0.63

2 0.5

3 0.37

4 0.13

Explanation:

D Given that,
Half-life of radioactive nuclei $(T_{1 / 2}) = 3$ days
Number of active nuclei left alter $t$ ,
$N = N_{0} {(\frac{1}{2})}^{n} (∵ n = \frac{t}{T_{1 / 2}})$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
After $t = 2$ days then-
$N_{1} = N_{0} {(\frac{1}{2})}^{\frac{2}{3}} = \frac{N_{0}}{2^{2 / 3}} = \frac{N_{0}}{4^{1 / 3}} = 0.63 N_{0}$
$N_{1} = 0.63 N_{0}$
After $t = 3$ days
$N_{2} = N_{0} {(\frac{1}{2})}^{\frac{3}{3}} = \frac{N_{0}}{2} = 0.5 N_{0}$
Number of nuclei that will decay on the $3^{rd}$ day,
$N_{3} = N_{1} - N_{2}$
$N_{3} = 0.63 N_{o} - 0.5 N_{o}$
$N_{3} = 0.13 N_{o}$

WB JEE 2018

NUCLEAR PHYSICS

147824 The half-life of tritium is $12.5$ years. What mass of tritium of initial mass $64 mg$ will remain undecayed after 50 years ?

1

32 mg

2

8 mg

3

16 mg

4

4 mg

Explanation:

D Given that,
Half-life of tritium $(T_{1 / 2}) = 12.5$ years
Initial mass $(N_{0}) = 64 mg$ .
Total time $(t) = 50$ year From the radioactive equation-
$N = N_{0} {(\frac{1}{2})}^{n}$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$N = 64 {(\frac{1}{2})}^{\frac{50}{12.5}} \Rightarrow N = 64 \times {(\frac{1}{2})}^{4}$
$N = 64 \times \frac{1}{16}$
$N = 4 mg$

Karnataka CET-2018

NUCLEAR PHYSICS

147825 The luminous dials watches are usually made by mixing a zinc sulphide phosphor with an aparticle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 a - particles per second for this purpose is

1

2.77 mg

2

2.77 g

3

2.77 \times 10^{- 23} g

4

2.77 \times 10^{- 13} kg

.

Explanation:

D Given that,
The rate of radioactive decay $= 10$
$\frac{dN}{dt} = 10$
For the first order decay -
$\frac{dN}{dt} = λ N$
$10 = λ N$
$N = \frac{10}{λ} (∵ λ = \frac{\ln 2}{t_{1 / 2}})$
$N = \frac{10}{\frac{\ln 2}{t_{1 / 2}}}$
$N = \frac{10 \times t_{1 / 2}}{\ln 2}$
$N = \frac{10 \times [1620 \times (365 \times 24 \times 3600)]}{N = 7.3 \times 10^{11}}$
$\ln 2$
Atom is present in,
$= \frac{N}{N_{A}} = \frac{7.3 \times 10^{11}}{6.023 \times 10^{23}} = 1.223 \times 10^{- 12} moles$
According to question -
Atomic mass of radium $= 226$ mole
1 Mole of radium as a mass of $226 g$
Mass of radium $= 1.223 \times 10^{- 12} \times 226 = 2.765 \times 10^{- 10} g$
$= 2.765 \times 10^{- 13} kg$

AIIMS-26.05.2018(M)

NUCLEAR PHYSICS

147820 A radioactive substance of half life 138.6 days is placed in a box. After $n$ days only $20 %$ of the substance is present then the value of $n$ is $[\ln (5) = 1.6]$

1 693

2 138.6

3 277.2

4 322

Explanation:

D Given that,
$t_{1 / 2} = 138.6 day$
$λ = \frac{0.693}{138.6} = 5 \times 10^{- 3} \sec$
$t = n day$
According to question -
$N = 20 % N_{0}$
$N = \frac{20}{100} \times N_{0}$
$N = \frac{N_{0}}{5}$
We know that, $N = N_{0} e^{- λ t}$
$\frac{N_{0}}{5} = N_{0} e^{- λ n}$
$\frac{1}{5} = \frac{1}{e^{λ n}}$
$e^{λ n} = 5$
$λ n = \ln 5$
$n = \frac{1.61}{5 \times 10^{- 3}} (∵ \ln 5 = 1.61)$
$n = 322 day$

AP EAMCET (22.04.2019) Shift-I

NUCLEAR PHYSICS

147821 An atomic power nuclear reactor can deliver $300 MW$ . The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $MeV$ . The number of uranium atoms fashioned per hour will be

1

30 \times 10^{25}

2

4 \times 10^{22}

3

10 \times 10^{20}

4

5 \times 10^{15}

Explanation:

B Given that,
Power $(P) = 300 MW$
Energy $(E) = 170 MeV$
We know that,
$P = n \frac{E}{T}$
$300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$300 \times 10^{6} = \frac{n \times 170 \times 1.6 \times 10^{- 13}}{t}$
So, the number of atom per second $= \frac{n}{t} = 1.102 \times 10^{19}$ Hence, the number of atom per hour $= 1.102 \times 10^{19} \times$ $3600 = 3.97 \times 10^{22}$
The number of atom per hours $\approx 4 \times 10^{22}$ .

JIPMER-2018

NUCLEAR PHYSICS

147823 If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day?
(Given, $\sqrt[3]{0.25} \approx 0.63$ )

1 0.63

2 0.5

3 0.37

4 0.13

Explanation:

D Given that,
Half-life of radioactive nuclei $(T_{1 / 2}) = 3$ days
Number of active nuclei left alter $t$ ,
$N = N_{0} {(\frac{1}{2})}^{n} (∵ n = \frac{t}{T_{1 / 2}})$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
After $t = 2$ days then-
$N_{1} = N_{0} {(\frac{1}{2})}^{\frac{2}{3}} = \frac{N_{0}}{2^{2 / 3}} = \frac{N_{0}}{4^{1 / 3}} = 0.63 N_{0}$
$N_{1} = 0.63 N_{0}$
After $t = 3$ days
$N_{2} = N_{0} {(\frac{1}{2})}^{\frac{3}{3}} = \frac{N_{0}}{2} = 0.5 N_{0}$
Number of nuclei that will decay on the $3^{rd}$ day,
$N_{3} = N_{1} - N_{2}$
$N_{3} = 0.63 N_{o} - 0.5 N_{o}$
$N_{3} = 0.13 N_{o}$

WB JEE 2018

NUCLEAR PHYSICS

147824 The half-life of tritium is $12.5$ years. What mass of tritium of initial mass $64 mg$ will remain undecayed after 50 years ?

1

32 mg

2

8 mg

3

16 mg

4

4 mg

Explanation:

D Given that,
Half-life of tritium $(T_{1 / 2}) = 12.5$ years
Initial mass $(N_{0}) = 64 mg$ .
Total time $(t) = 50$ year From the radioactive equation-
$N = N_{0} {(\frac{1}{2})}^{n}$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$N = 64 {(\frac{1}{2})}^{\frac{50}{12.5}} \Rightarrow N = 64 \times {(\frac{1}{2})}^{4}$
$N = 64 \times \frac{1}{16}$
$N = 4 mg$

Karnataka CET-2018

NUCLEAR PHYSICS

147825 The luminous dials watches are usually made by mixing a zinc sulphide phosphor with an aparticle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 a - particles per second for this purpose is

1

2.77 mg

2

2.77 g

3

2.77 \times 10^{- 23} g

4

2.77 \times 10^{- 13} kg

.

Explanation:

D Given that,
The rate of radioactive decay $= 10$
$\frac{dN}{dt} = 10$
For the first order decay -
$\frac{dN}{dt} = λ N$
$10 = λ N$
$N = \frac{10}{λ} (∵ λ = \frac{\ln 2}{t_{1 / 2}})$
$N = \frac{10}{\frac{\ln 2}{t_{1 / 2}}}$
$N = \frac{10 \times t_{1 / 2}}{\ln 2}$
$N = \frac{10 \times [1620 \times (365 \times 24 \times 3600)]}{N = 7.3 \times 10^{11}}$
$\ln 2$
Atom is present in,
$= \frac{N}{N_{A}} = \frac{7.3 \times 10^{11}}{6.023 \times 10^{23}} = 1.223 \times 10^{- 12} moles$
According to question -
Atomic mass of radium $= 226$ mole
1 Mole of radium as a mass of $226 g$
Mass of radium $= 1.223 \times 10^{- 12} \times 226 = 2.765 \times 10^{- 10} g$
$= 2.765 \times 10^{- 13} kg$

AIIMS-26.05.2018(M)

NUCLEAR PHYSICS

147820 A radioactive substance of half life 138.6 days is placed in a box. After $n$ days only $20 %$ of the substance is present then the value of $n$ is $[\ln (5) = 1.6]$

1 693

2 138.6

3 277.2

4 322

Explanation:

D Given that,
$t_{1 / 2} = 138.6 day$
$λ = \frac{0.693}{138.6} = 5 \times 10^{- 3} \sec$
$t = n day$
According to question -
$N = 20 % N_{0}$
$N = \frac{20}{100} \times N_{0}$
$N = \frac{N_{0}}{5}$
We know that, $N = N_{0} e^{- λ t}$
$\frac{N_{0}}{5} = N_{0} e^{- λ n}$
$\frac{1}{5} = \frac{1}{e^{λ n}}$
$e^{λ n} = 5$
$λ n = \ln 5$
$n = \frac{1.61}{5 \times 10^{- 3}} (∵ \ln 5 = 1.61)$
$n = 322 day$

AP EAMCET (22.04.2019) Shift-I

NUCLEAR PHYSICS

147821 An atomic power nuclear reactor can deliver $300 MW$ . The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $MeV$ . The number of uranium atoms fashioned per hour will be

1

30 \times 10^{25}

2

4 \times 10^{22}

3

10 \times 10^{20}

4

5 \times 10^{15}

Explanation:

B Given that,
Power $(P) = 300 MW$
Energy $(E) = 170 MeV$
We know that,
$P = n \frac{E}{T}$
$300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$300 \times 10^{6} = \frac{n \times 170 \times 1.6 \times 10^{- 13}}{t}$
So, the number of atom per second $= \frac{n}{t} = 1.102 \times 10^{19}$ Hence, the number of atom per hour $= 1.102 \times 10^{19} \times$ $3600 = 3.97 \times 10^{22}$
The number of atom per hours $\approx 4 \times 10^{22}$ .

JIPMER-2018

NUCLEAR PHYSICS

147823 If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day?
(Given, $\sqrt[3]{0.25} \approx 0.63$ )

1 0.63

2 0.5

3 0.37

4 0.13

Explanation:

D Given that,
Half-life of radioactive nuclei $(T_{1 / 2}) = 3$ days
Number of active nuclei left alter $t$ ,
$N = N_{0} {(\frac{1}{2})}^{n} (∵ n = \frac{t}{T_{1 / 2}})$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
After $t = 2$ days then-
$N_{1} = N_{0} {(\frac{1}{2})}^{\frac{2}{3}} = \frac{N_{0}}{2^{2 / 3}} = \frac{N_{0}}{4^{1 / 3}} = 0.63 N_{0}$
$N_{1} = 0.63 N_{0}$
After $t = 3$ days
$N_{2} = N_{0} {(\frac{1}{2})}^{\frac{3}{3}} = \frac{N_{0}}{2} = 0.5 N_{0}$
Number of nuclei that will decay on the $3^{rd}$ day,
$N_{3} = N_{1} - N_{2}$
$N_{3} = 0.63 N_{o} - 0.5 N_{o}$
$N_{3} = 0.13 N_{o}$

WB JEE 2018

NUCLEAR PHYSICS

147824 The half-life of tritium is $12.5$ years. What mass of tritium of initial mass $64 mg$ will remain undecayed after 50 years ?

1

32 mg

2

8 mg

3

16 mg

4

4 mg

Explanation:

D Given that,
Half-life of tritium $(T_{1 / 2}) = 12.5$ years
Initial mass $(N_{0}) = 64 mg$ .
Total time $(t) = 50$ year From the radioactive equation-
$N = N_{0} {(\frac{1}{2})}^{n}$
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$N = 64 {(\frac{1}{2})}^{\frac{50}{12.5}} \Rightarrow N = 64 \times {(\frac{1}{2})}^{4}$
$N = 64 \times \frac{1}{16}$
$N = 4 mg$

Karnataka CET-2018

NUCLEAR PHYSICS

147825 The luminous dials watches are usually made by mixing a zinc sulphide phosphor with an aparticle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 a - particles per second for this purpose is

1

2.77 mg

2

2.77 g

3

2.77 \times 10^{- 23} g

4

2.77 \times 10^{- 13} kg

.

Explanation:

D Given that,
The rate of radioactive decay $= 10$
$\frac{dN}{dt} = 10$
For the first order decay -
$\frac{dN}{dt} = λ N$
$10 = λ N$
$N = \frac{10}{λ} (∵ λ = \frac{\ln 2}{t_{1 / 2}})$
$N = \frac{10}{\frac{\ln 2}{t_{1 / 2}}}$
$N = \frac{10 \times t_{1 / 2}}{\ln 2}$
$N = \frac{10 \times [1620 \times (365 \times 24 \times 3600)]}{N = 7.3 \times 10^{11}}$
$\ln 2$
Atom is present in,
$= \frac{N}{N_{A}} = \frac{7.3 \times 10^{11}}{6.023 \times 10^{23}} = 1.223 \times 10^{- 12} moles$
According to question -
Atomic mass of radium $= 226$ mole
1 Mole of radium as a mass of $226 g$
Mass of radium $= 1.223 \times 10^{- 12} \times 226 = 2.765 \times 10^{- 10} g$
$= 2.765 \times 10^{- 13} kg$

AIIMS-26.05.2018(M)