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, $\mathrm{t}_{1 / 2}=138.6 \text { day }$ $\lambda=\frac{0.693}{138.6}=5 \times 10^{-3} \mathrm{sec}$ $\mathrm{t}=\mathrm{n} \text { day }$ According to question - $\mathrm{N}=20 \% \mathrm{~N}_{0}$ $\mathrm{~N}=\frac{20}{100} \times \mathrm{N}_{0}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{5}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{0}}{5}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{n}}$ $\frac{1}{5}=\frac{1}{\mathrm{e}^{\lambda \mathrm{n}}}$ $\mathrm{e}^{\lambda \mathrm{n}}=5$ $\lambda \mathrm{n}=\ln 5$ $\mathrm{n}=\frac{1.61}{5 \times 10^{-3}} \quad(\because \ln 5=1.61)$ $\mathrm{n}=322 \text { day }$
AP EAMCET (22.04.2019) Shift-I
NUCLEAR PHYSICS
147821
An atomic power nuclear reactor can deliver $300 \mathrm{MW}$. The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $\mathrm{MeV}$. 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 $(\mathrm{P})=300 \mathrm{MW}$ Energy $(\mathrm{E})=170 \mathrm{MeV}$ 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 $=\frac{\mathrm{n}}{\mathrm{t}}=1.102 \times 10^{19}$ 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, $\sqrt[3]{\mathbf{0 . 2 5}} \approx 0.63$ )
1 0.63
2 0.5
3 0.37
4 0.13
Explanation:
D Given that, Half-life of radioactive nuclei $\left(\mathrm{T}_{1 / 2}\right)=3$ days Number of active nuclei left alter $t$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ After $\mathrm{t}=2$ days then- $\mathrm{N}_{1}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{2}{3}}=\frac{\mathrm{N}_{0}}{2^{2 / 3}}=\frac{\mathrm{N}_{0}}{4^{1 / 3}}=0.63 \mathrm{~N}_{0}$ $\mathrm{~N}_{1}=0.63 \mathrm{~N}_{0}$ After $\mathrm{t}=3$ days $\mathrm{N}_{2}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{3}{3}}=\frac{\mathrm{N}_{0}}{2}=0.5 \mathrm{~N}_{0}$ Number of nuclei that will decay on the $3^{\text {rd }}$ day, $\mathrm{N}_{3}=\mathrm{N}_{1}-\mathrm{N}_{2}$ $\mathrm{~N}_{3}=0.63 \mathrm{~N}_{\mathrm{o}}-0.5 \mathrm{~N}_{\mathrm{o}}$ $\mathrm{N}_{3}=0.13 \mathrm{~N}_{\mathrm{o}}$
WB JEE 2018
NUCLEAR PHYSICS
147824
The half-life of tritium is $\mathbf{1 2 . 5}$ years. What mass of tritium of initial mass $64 \mathrm{mg}$ will remain undecayed after 50 years ?
1 $32 \mathrm{mg}$
2 $8 \mathrm{mg}$
3 $16 \mathrm{mg}$
4 $4 \mathrm{mg}$
Explanation:
D Given that, Half-life of tritium $\left(\mathrm{T}_{1 / 2}\right)=12.5$ years Initial mass $\left(\mathrm{N}_{0}\right)=64 \mathrm{mg}$. Total time $(\mathrm{t})=50$ year From the radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=64\left(\frac{1}{2}\right)^{\frac{50}{12.5}} \Rightarrow \quad \mathrm{N}=64 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=64 \times \frac{1}{16}$ $\mathrm{~N}=4 \mathrm{mg}$
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 $2.77 \mathrm{mg}$
2 $2.77 \mathrm{~g}$
3 $2.77 \times 10^{-23} \mathrm{~g}$
4 $2.77 \times 10^{-13} \mathrm{~kg}$.
Explanation:
D Given that, The rate of radioactive decay $=10$ $\frac{\mathrm{dN}}{\mathrm{dt}}=10$ For the first order decay - $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $10=\lambda \mathrm{N}$ $\mathrm{N}=\frac{10}{\lambda} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right)$ $\mathrm{N}=\frac{10}{\frac{\ln 2}{\mathrm{t}_{1 / 2}}}$ $\mathrm{~N}=\frac{10 \times \mathrm{t}_{1 / 2}}{\ln 2}$ $\mathrm{~N}=\frac{10 \times[1620 \times(365 \times 24 \times 3600)]}{\mathrm{N}=7.3 \times 10^{11}}$ $\ln 2$ Atom is present in, $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}=\frac{7.3 \times 10^{11}}{6.023 \times 10^{23}}=1.223 \times 10^{-12} \mathrm{moles}$ According to question - Atomic mass of radium $=226$ mole 1 Mole of radium as a mass of $226 \mathrm{~g}$ Mass of radium $=1.223 \times 10^{-12} \times 226=2.765 \times 10^{-10} \mathrm{~g}$ $=2.765 \times 10^{-13} \mathrm{~kg}$
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, $\mathrm{t}_{1 / 2}=138.6 \text { day }$ $\lambda=\frac{0.693}{138.6}=5 \times 10^{-3} \mathrm{sec}$ $\mathrm{t}=\mathrm{n} \text { day }$ According to question - $\mathrm{N}=20 \% \mathrm{~N}_{0}$ $\mathrm{~N}=\frac{20}{100} \times \mathrm{N}_{0}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{5}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{0}}{5}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{n}}$ $\frac{1}{5}=\frac{1}{\mathrm{e}^{\lambda \mathrm{n}}}$ $\mathrm{e}^{\lambda \mathrm{n}}=5$ $\lambda \mathrm{n}=\ln 5$ $\mathrm{n}=\frac{1.61}{5 \times 10^{-3}} \quad(\because \ln 5=1.61)$ $\mathrm{n}=322 \text { day }$
AP EAMCET (22.04.2019) Shift-I
NUCLEAR PHYSICS
147821
An atomic power nuclear reactor can deliver $300 \mathrm{MW}$. The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $\mathrm{MeV}$. 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 $(\mathrm{P})=300 \mathrm{MW}$ Energy $(\mathrm{E})=170 \mathrm{MeV}$ 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 $=\frac{\mathrm{n}}{\mathrm{t}}=1.102 \times 10^{19}$ 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, $\sqrt[3]{\mathbf{0 . 2 5}} \approx 0.63$ )
1 0.63
2 0.5
3 0.37
4 0.13
Explanation:
D Given that, Half-life of radioactive nuclei $\left(\mathrm{T}_{1 / 2}\right)=3$ days Number of active nuclei left alter $t$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ After $\mathrm{t}=2$ days then- $\mathrm{N}_{1}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{2}{3}}=\frac{\mathrm{N}_{0}}{2^{2 / 3}}=\frac{\mathrm{N}_{0}}{4^{1 / 3}}=0.63 \mathrm{~N}_{0}$ $\mathrm{~N}_{1}=0.63 \mathrm{~N}_{0}$ After $\mathrm{t}=3$ days $\mathrm{N}_{2}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{3}{3}}=\frac{\mathrm{N}_{0}}{2}=0.5 \mathrm{~N}_{0}$ Number of nuclei that will decay on the $3^{\text {rd }}$ day, $\mathrm{N}_{3}=\mathrm{N}_{1}-\mathrm{N}_{2}$ $\mathrm{~N}_{3}=0.63 \mathrm{~N}_{\mathrm{o}}-0.5 \mathrm{~N}_{\mathrm{o}}$ $\mathrm{N}_{3}=0.13 \mathrm{~N}_{\mathrm{o}}$
WB JEE 2018
NUCLEAR PHYSICS
147824
The half-life of tritium is $\mathbf{1 2 . 5}$ years. What mass of tritium of initial mass $64 \mathrm{mg}$ will remain undecayed after 50 years ?
1 $32 \mathrm{mg}$
2 $8 \mathrm{mg}$
3 $16 \mathrm{mg}$
4 $4 \mathrm{mg}$
Explanation:
D Given that, Half-life of tritium $\left(\mathrm{T}_{1 / 2}\right)=12.5$ years Initial mass $\left(\mathrm{N}_{0}\right)=64 \mathrm{mg}$. Total time $(\mathrm{t})=50$ year From the radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=64\left(\frac{1}{2}\right)^{\frac{50}{12.5}} \Rightarrow \quad \mathrm{N}=64 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=64 \times \frac{1}{16}$ $\mathrm{~N}=4 \mathrm{mg}$
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 $2.77 \mathrm{mg}$
2 $2.77 \mathrm{~g}$
3 $2.77 \times 10^{-23} \mathrm{~g}$
4 $2.77 \times 10^{-13} \mathrm{~kg}$.
Explanation:
D Given that, The rate of radioactive decay $=10$ $\frac{\mathrm{dN}}{\mathrm{dt}}=10$ For the first order decay - $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $10=\lambda \mathrm{N}$ $\mathrm{N}=\frac{10}{\lambda} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right)$ $\mathrm{N}=\frac{10}{\frac{\ln 2}{\mathrm{t}_{1 / 2}}}$ $\mathrm{~N}=\frac{10 \times \mathrm{t}_{1 / 2}}{\ln 2}$ $\mathrm{~N}=\frac{10 \times[1620 \times(365 \times 24 \times 3600)]}{\mathrm{N}=7.3 \times 10^{11}}$ $\ln 2$ Atom is present in, $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}=\frac{7.3 \times 10^{11}}{6.023 \times 10^{23}}=1.223 \times 10^{-12} \mathrm{moles}$ According to question - Atomic mass of radium $=226$ mole 1 Mole of radium as a mass of $226 \mathrm{~g}$ Mass of radium $=1.223 \times 10^{-12} \times 226=2.765 \times 10^{-10} \mathrm{~g}$ $=2.765 \times 10^{-13} \mathrm{~kg}$
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, $\mathrm{t}_{1 / 2}=138.6 \text { day }$ $\lambda=\frac{0.693}{138.6}=5 \times 10^{-3} \mathrm{sec}$ $\mathrm{t}=\mathrm{n} \text { day }$ According to question - $\mathrm{N}=20 \% \mathrm{~N}_{0}$ $\mathrm{~N}=\frac{20}{100} \times \mathrm{N}_{0}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{5}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{0}}{5}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{n}}$ $\frac{1}{5}=\frac{1}{\mathrm{e}^{\lambda \mathrm{n}}}$ $\mathrm{e}^{\lambda \mathrm{n}}=5$ $\lambda \mathrm{n}=\ln 5$ $\mathrm{n}=\frac{1.61}{5 \times 10^{-3}} \quad(\because \ln 5=1.61)$ $\mathrm{n}=322 \text { day }$
AP EAMCET (22.04.2019) Shift-I
NUCLEAR PHYSICS
147821
An atomic power nuclear reactor can deliver $300 \mathrm{MW}$. The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $\mathrm{MeV}$. 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 $(\mathrm{P})=300 \mathrm{MW}$ Energy $(\mathrm{E})=170 \mathrm{MeV}$ 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 $=\frac{\mathrm{n}}{\mathrm{t}}=1.102 \times 10^{19}$ 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, $\sqrt[3]{\mathbf{0 . 2 5}} \approx 0.63$ )
1 0.63
2 0.5
3 0.37
4 0.13
Explanation:
D Given that, Half-life of radioactive nuclei $\left(\mathrm{T}_{1 / 2}\right)=3$ days Number of active nuclei left alter $t$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ After $\mathrm{t}=2$ days then- $\mathrm{N}_{1}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{2}{3}}=\frac{\mathrm{N}_{0}}{2^{2 / 3}}=\frac{\mathrm{N}_{0}}{4^{1 / 3}}=0.63 \mathrm{~N}_{0}$ $\mathrm{~N}_{1}=0.63 \mathrm{~N}_{0}$ After $\mathrm{t}=3$ days $\mathrm{N}_{2}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{3}{3}}=\frac{\mathrm{N}_{0}}{2}=0.5 \mathrm{~N}_{0}$ Number of nuclei that will decay on the $3^{\text {rd }}$ day, $\mathrm{N}_{3}=\mathrm{N}_{1}-\mathrm{N}_{2}$ $\mathrm{~N}_{3}=0.63 \mathrm{~N}_{\mathrm{o}}-0.5 \mathrm{~N}_{\mathrm{o}}$ $\mathrm{N}_{3}=0.13 \mathrm{~N}_{\mathrm{o}}$
WB JEE 2018
NUCLEAR PHYSICS
147824
The half-life of tritium is $\mathbf{1 2 . 5}$ years. What mass of tritium of initial mass $64 \mathrm{mg}$ will remain undecayed after 50 years ?
1 $32 \mathrm{mg}$
2 $8 \mathrm{mg}$
3 $16 \mathrm{mg}$
4 $4 \mathrm{mg}$
Explanation:
D Given that, Half-life of tritium $\left(\mathrm{T}_{1 / 2}\right)=12.5$ years Initial mass $\left(\mathrm{N}_{0}\right)=64 \mathrm{mg}$. Total time $(\mathrm{t})=50$ year From the radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=64\left(\frac{1}{2}\right)^{\frac{50}{12.5}} \Rightarrow \quad \mathrm{N}=64 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=64 \times \frac{1}{16}$ $\mathrm{~N}=4 \mathrm{mg}$
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 $2.77 \mathrm{mg}$
2 $2.77 \mathrm{~g}$
3 $2.77 \times 10^{-23} \mathrm{~g}$
4 $2.77 \times 10^{-13} \mathrm{~kg}$.
Explanation:
D Given that, The rate of radioactive decay $=10$ $\frac{\mathrm{dN}}{\mathrm{dt}}=10$ For the first order decay - $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $10=\lambda \mathrm{N}$ $\mathrm{N}=\frac{10}{\lambda} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right)$ $\mathrm{N}=\frac{10}{\frac{\ln 2}{\mathrm{t}_{1 / 2}}}$ $\mathrm{~N}=\frac{10 \times \mathrm{t}_{1 / 2}}{\ln 2}$ $\mathrm{~N}=\frac{10 \times[1620 \times(365 \times 24 \times 3600)]}{\mathrm{N}=7.3 \times 10^{11}}$ $\ln 2$ Atom is present in, $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}=\frac{7.3 \times 10^{11}}{6.023 \times 10^{23}}=1.223 \times 10^{-12} \mathrm{moles}$ According to question - Atomic mass of radium $=226$ mole 1 Mole of radium as a mass of $226 \mathrm{~g}$ Mass of radium $=1.223 \times 10^{-12} \times 226=2.765 \times 10^{-10} \mathrm{~g}$ $=2.765 \times 10^{-13} \mathrm{~kg}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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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, $\mathrm{t}_{1 / 2}=138.6 \text { day }$ $\lambda=\frac{0.693}{138.6}=5 \times 10^{-3} \mathrm{sec}$ $\mathrm{t}=\mathrm{n} \text { day }$ According to question - $\mathrm{N}=20 \% \mathrm{~N}_{0}$ $\mathrm{~N}=\frac{20}{100} \times \mathrm{N}_{0}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{5}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{0}}{5}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{n}}$ $\frac{1}{5}=\frac{1}{\mathrm{e}^{\lambda \mathrm{n}}}$ $\mathrm{e}^{\lambda \mathrm{n}}=5$ $\lambda \mathrm{n}=\ln 5$ $\mathrm{n}=\frac{1.61}{5 \times 10^{-3}} \quad(\because \ln 5=1.61)$ $\mathrm{n}=322 \text { day }$
AP EAMCET (22.04.2019) Shift-I
NUCLEAR PHYSICS
147821
An atomic power nuclear reactor can deliver $300 \mathrm{MW}$. The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $\mathrm{MeV}$. 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 $(\mathrm{P})=300 \mathrm{MW}$ Energy $(\mathrm{E})=170 \mathrm{MeV}$ 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 $=\frac{\mathrm{n}}{\mathrm{t}}=1.102 \times 10^{19}$ 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, $\sqrt[3]{\mathbf{0 . 2 5}} \approx 0.63$ )
1 0.63
2 0.5
3 0.37
4 0.13
Explanation:
D Given that, Half-life of radioactive nuclei $\left(\mathrm{T}_{1 / 2}\right)=3$ days Number of active nuclei left alter $t$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ After $\mathrm{t}=2$ days then- $\mathrm{N}_{1}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{2}{3}}=\frac{\mathrm{N}_{0}}{2^{2 / 3}}=\frac{\mathrm{N}_{0}}{4^{1 / 3}}=0.63 \mathrm{~N}_{0}$ $\mathrm{~N}_{1}=0.63 \mathrm{~N}_{0}$ After $\mathrm{t}=3$ days $\mathrm{N}_{2}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{3}{3}}=\frac{\mathrm{N}_{0}}{2}=0.5 \mathrm{~N}_{0}$ Number of nuclei that will decay on the $3^{\text {rd }}$ day, $\mathrm{N}_{3}=\mathrm{N}_{1}-\mathrm{N}_{2}$ $\mathrm{~N}_{3}=0.63 \mathrm{~N}_{\mathrm{o}}-0.5 \mathrm{~N}_{\mathrm{o}}$ $\mathrm{N}_{3}=0.13 \mathrm{~N}_{\mathrm{o}}$
WB JEE 2018
NUCLEAR PHYSICS
147824
The half-life of tritium is $\mathbf{1 2 . 5}$ years. What mass of tritium of initial mass $64 \mathrm{mg}$ will remain undecayed after 50 years ?
1 $32 \mathrm{mg}$
2 $8 \mathrm{mg}$
3 $16 \mathrm{mg}$
4 $4 \mathrm{mg}$
Explanation:
D Given that, Half-life of tritium $\left(\mathrm{T}_{1 / 2}\right)=12.5$ years Initial mass $\left(\mathrm{N}_{0}\right)=64 \mathrm{mg}$. Total time $(\mathrm{t})=50$ year From the radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=64\left(\frac{1}{2}\right)^{\frac{50}{12.5}} \Rightarrow \quad \mathrm{N}=64 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=64 \times \frac{1}{16}$ $\mathrm{~N}=4 \mathrm{mg}$
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 $2.77 \mathrm{mg}$
2 $2.77 \mathrm{~g}$
3 $2.77 \times 10^{-23} \mathrm{~g}$
4 $2.77 \times 10^{-13} \mathrm{~kg}$.
Explanation:
D Given that, The rate of radioactive decay $=10$ $\frac{\mathrm{dN}}{\mathrm{dt}}=10$ For the first order decay - $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $10=\lambda \mathrm{N}$ $\mathrm{N}=\frac{10}{\lambda} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right)$ $\mathrm{N}=\frac{10}{\frac{\ln 2}{\mathrm{t}_{1 / 2}}}$ $\mathrm{~N}=\frac{10 \times \mathrm{t}_{1 / 2}}{\ln 2}$ $\mathrm{~N}=\frac{10 \times[1620 \times(365 \times 24 \times 3600)]}{\mathrm{N}=7.3 \times 10^{11}}$ $\ln 2$ Atom is present in, $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}=\frac{7.3 \times 10^{11}}{6.023 \times 10^{23}}=1.223 \times 10^{-12} \mathrm{moles}$ According to question - Atomic mass of radium $=226$ mole 1 Mole of radium as a mass of $226 \mathrm{~g}$ Mass of radium $=1.223 \times 10^{-12} \times 226=2.765 \times 10^{-10} \mathrm{~g}$ $=2.765 \times 10^{-13} \mathrm{~kg}$
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, $\mathrm{t}_{1 / 2}=138.6 \text { day }$ $\lambda=\frac{0.693}{138.6}=5 \times 10^{-3} \mathrm{sec}$ $\mathrm{t}=\mathrm{n} \text { day }$ According to question - $\mathrm{N}=20 \% \mathrm{~N}_{0}$ $\mathrm{~N}=\frac{20}{100} \times \mathrm{N}_{0}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{5}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{0}}{5}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{n}}$ $\frac{1}{5}=\frac{1}{\mathrm{e}^{\lambda \mathrm{n}}}$ $\mathrm{e}^{\lambda \mathrm{n}}=5$ $\lambda \mathrm{n}=\ln 5$ $\mathrm{n}=\frac{1.61}{5 \times 10^{-3}} \quad(\because \ln 5=1.61)$ $\mathrm{n}=322 \text { day }$
AP EAMCET (22.04.2019) Shift-I
NUCLEAR PHYSICS
147821
An atomic power nuclear reactor can deliver $300 \mathrm{MW}$. The energy released due to fission of each nucleus of uranium atoms $U^{238}$ is 170 $\mathrm{MeV}$. 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 $(\mathrm{P})=300 \mathrm{MW}$ Energy $(\mathrm{E})=170 \mathrm{MeV}$ 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 $=\frac{\mathrm{n}}{\mathrm{t}}=1.102 \times 10^{19}$ 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, $\sqrt[3]{\mathbf{0 . 2 5}} \approx 0.63$ )
1 0.63
2 0.5
3 0.37
4 0.13
Explanation:
D Given that, Half-life of radioactive nuclei $\left(\mathrm{T}_{1 / 2}\right)=3$ days Number of active nuclei left alter $t$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ After $\mathrm{t}=2$ days then- $\mathrm{N}_{1}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{2}{3}}=\frac{\mathrm{N}_{0}}{2^{2 / 3}}=\frac{\mathrm{N}_{0}}{4^{1 / 3}}=0.63 \mathrm{~N}_{0}$ $\mathrm{~N}_{1}=0.63 \mathrm{~N}_{0}$ After $\mathrm{t}=3$ days $\mathrm{N}_{2}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{3}{3}}=\frac{\mathrm{N}_{0}}{2}=0.5 \mathrm{~N}_{0}$ Number of nuclei that will decay on the $3^{\text {rd }}$ day, $\mathrm{N}_{3}=\mathrm{N}_{1}-\mathrm{N}_{2}$ $\mathrm{~N}_{3}=0.63 \mathrm{~N}_{\mathrm{o}}-0.5 \mathrm{~N}_{\mathrm{o}}$ $\mathrm{N}_{3}=0.13 \mathrm{~N}_{\mathrm{o}}$
WB JEE 2018
NUCLEAR PHYSICS
147824
The half-life of tritium is $\mathbf{1 2 . 5}$ years. What mass of tritium of initial mass $64 \mathrm{mg}$ will remain undecayed after 50 years ?
1 $32 \mathrm{mg}$
2 $8 \mathrm{mg}$
3 $16 \mathrm{mg}$
4 $4 \mathrm{mg}$
Explanation:
D Given that, Half-life of tritium $\left(\mathrm{T}_{1 / 2}\right)=12.5$ years Initial mass $\left(\mathrm{N}_{0}\right)=64 \mathrm{mg}$. Total time $(\mathrm{t})=50$ year From the radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=64\left(\frac{1}{2}\right)^{\frac{50}{12.5}} \Rightarrow \quad \mathrm{N}=64 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=64 \times \frac{1}{16}$ $\mathrm{~N}=4 \mathrm{mg}$
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 $2.77 \mathrm{mg}$
2 $2.77 \mathrm{~g}$
3 $2.77 \times 10^{-23} \mathrm{~g}$
4 $2.77 \times 10^{-13} \mathrm{~kg}$.
Explanation:
D Given that, The rate of radioactive decay $=10$ $\frac{\mathrm{dN}}{\mathrm{dt}}=10$ For the first order decay - $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $10=\lambda \mathrm{N}$ $\mathrm{N}=\frac{10}{\lambda} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right)$ $\mathrm{N}=\frac{10}{\frac{\ln 2}{\mathrm{t}_{1 / 2}}}$ $\mathrm{~N}=\frac{10 \times \mathrm{t}_{1 / 2}}{\ln 2}$ $\mathrm{~N}=\frac{10 \times[1620 \times(365 \times 24 \times 3600)]}{\mathrm{N}=7.3 \times 10^{11}}$ $\ln 2$ Atom is present in, $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}=\frac{7.3 \times 10^{11}}{6.023 \times 10^{23}}=1.223 \times 10^{-12} \mathrm{moles}$ According to question - Atomic mass of radium $=226$ mole 1 Mole of radium as a mass of $226 \mathrm{~g}$ Mass of radium $=1.223 \times 10^{-12} \times 226=2.765 \times 10^{-10} \mathrm{~g}$ $=2.765 \times 10^{-13} \mathrm{~kg}$