148008
Calculate power output of ${ }_{92}^{235} U$ reactor, if it takes 30 days to use up $2 \mathrm{~kg}$ of fuel, and if each fission gives $185 \mathrm{MeV}$ of useable energy. A Avogadro's number $=6 \times 10^{23} / \mathrm{mol}$ ?
148011
A radioactive substance decays to $1 / 16^{\text {th }}$ of its initial activity in 40 days. The half-life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given that, Total time taken in decay $(\mathrm{t})=40$ days According to question - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ On comparing both the side, we get- $\frac{40}{\mathrm{~T}_{1 / 2}}=4$ $\mathrm{~T}_{1 / 2}=\frac{40}{4}$ $\mathrm{~T}_{1 / 2}=10 \text { days }$
AIIMS-2003
NUCLEAR PHYSICS
148015
The following equation represents induced transmutation ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\mathrm{X}$ In this equation, $X$ represents
1 one $\beta^{-}$particle
2 $\alpha$-particle
3 a positron
4 a neutron
Explanation:
D Given reaction, ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ Sum of atomic number in L.H.S $=4+2=6$ Sum of atomic number in R.H.S $=6+Z$ Then, $\text { L.H.S = R.H.S }$ $4+2=6+Z$ $Z=0$ Sum of mass number in L.H.S $=9+4=13$ Sum of mass number in R.H.S $=12+$ A Then, $13=12+A$ $A=13-12$ $A=1$ So, the emitted particle is neutron $\left({ }_{0} \mathrm{n}^{1}\right)$.
BCECE-2010
NUCLEAR PHYSICS
148016
A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866 \mathrm{u}$. The energy liberated per $u$ is (given $1 \mathrm{u} 931 \mathrm{MeV}$ )
1 $2.67 \mathrm{MeV}$
2 $26.7 \mathrm{MeV}$
3 $6.675 \mathrm{MeV}$
4 $13.35 \mathrm{MeV}$
Explanation:
C Given that, Mass of reaction $(\Delta \mathrm{m})=0.02866 \mathrm{u}$ $1 \mathrm{u}=931 \mathrm{MeV}$ The nuclear fusion reaction, ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$ Energy released $\left(\mathrm{E}_{\mathrm{T}}\right)=\Delta \mathrm{m} \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \mathrm{u} \times \frac{931 \frac{\mathrm{MeV}}{\mathrm{c}^{2}}}{\mathrm{u}} \times \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \times 931.5$ $\mathrm{E}_{\mathrm{T}}=26.697$ We know that atomic mass of helium is 4 . So, Total energy released per $\mathrm{u}$ is - $\mathrm{E}=\frac{\mathrm{E}_{\mathrm{T}}}{4}$ $\mathrm{E}=\frac{26.697}{4}$ $\mathrm{E}=6.675 \mathrm{MeV}$
148008
Calculate power output of ${ }_{92}^{235} U$ reactor, if it takes 30 days to use up $2 \mathrm{~kg}$ of fuel, and if each fission gives $185 \mathrm{MeV}$ of useable energy. A Avogadro's number $=6 \times 10^{23} / \mathrm{mol}$ ?
148011
A radioactive substance decays to $1 / 16^{\text {th }}$ of its initial activity in 40 days. The half-life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given that, Total time taken in decay $(\mathrm{t})=40$ days According to question - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ On comparing both the side, we get- $\frac{40}{\mathrm{~T}_{1 / 2}}=4$ $\mathrm{~T}_{1 / 2}=\frac{40}{4}$ $\mathrm{~T}_{1 / 2}=10 \text { days }$
AIIMS-2003
NUCLEAR PHYSICS
148015
The following equation represents induced transmutation ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\mathrm{X}$ In this equation, $X$ represents
1 one $\beta^{-}$particle
2 $\alpha$-particle
3 a positron
4 a neutron
Explanation:
D Given reaction, ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ Sum of atomic number in L.H.S $=4+2=6$ Sum of atomic number in R.H.S $=6+Z$ Then, $\text { L.H.S = R.H.S }$ $4+2=6+Z$ $Z=0$ Sum of mass number in L.H.S $=9+4=13$ Sum of mass number in R.H.S $=12+$ A Then, $13=12+A$ $A=13-12$ $A=1$ So, the emitted particle is neutron $\left({ }_{0} \mathrm{n}^{1}\right)$.
BCECE-2010
NUCLEAR PHYSICS
148016
A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866 \mathrm{u}$. The energy liberated per $u$ is (given $1 \mathrm{u} 931 \mathrm{MeV}$ )
1 $2.67 \mathrm{MeV}$
2 $26.7 \mathrm{MeV}$
3 $6.675 \mathrm{MeV}$
4 $13.35 \mathrm{MeV}$
Explanation:
C Given that, Mass of reaction $(\Delta \mathrm{m})=0.02866 \mathrm{u}$ $1 \mathrm{u}=931 \mathrm{MeV}$ The nuclear fusion reaction, ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$ Energy released $\left(\mathrm{E}_{\mathrm{T}}\right)=\Delta \mathrm{m} \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \mathrm{u} \times \frac{931 \frac{\mathrm{MeV}}{\mathrm{c}^{2}}}{\mathrm{u}} \times \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \times 931.5$ $\mathrm{E}_{\mathrm{T}}=26.697$ We know that atomic mass of helium is 4 . So, Total energy released per $\mathrm{u}$ is - $\mathrm{E}=\frac{\mathrm{E}_{\mathrm{T}}}{4}$ $\mathrm{E}=\frac{26.697}{4}$ $\mathrm{E}=6.675 \mathrm{MeV}$
148008
Calculate power output of ${ }_{92}^{235} U$ reactor, if it takes 30 days to use up $2 \mathrm{~kg}$ of fuel, and if each fission gives $185 \mathrm{MeV}$ of useable energy. A Avogadro's number $=6 \times 10^{23} / \mathrm{mol}$ ?
148011
A radioactive substance decays to $1 / 16^{\text {th }}$ of its initial activity in 40 days. The half-life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given that, Total time taken in decay $(\mathrm{t})=40$ days According to question - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ On comparing both the side, we get- $\frac{40}{\mathrm{~T}_{1 / 2}}=4$ $\mathrm{~T}_{1 / 2}=\frac{40}{4}$ $\mathrm{~T}_{1 / 2}=10 \text { days }$
AIIMS-2003
NUCLEAR PHYSICS
148015
The following equation represents induced transmutation ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\mathrm{X}$ In this equation, $X$ represents
1 one $\beta^{-}$particle
2 $\alpha$-particle
3 a positron
4 a neutron
Explanation:
D Given reaction, ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ Sum of atomic number in L.H.S $=4+2=6$ Sum of atomic number in R.H.S $=6+Z$ Then, $\text { L.H.S = R.H.S }$ $4+2=6+Z$ $Z=0$ Sum of mass number in L.H.S $=9+4=13$ Sum of mass number in R.H.S $=12+$ A Then, $13=12+A$ $A=13-12$ $A=1$ So, the emitted particle is neutron $\left({ }_{0} \mathrm{n}^{1}\right)$.
BCECE-2010
NUCLEAR PHYSICS
148016
A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866 \mathrm{u}$. The energy liberated per $u$ is (given $1 \mathrm{u} 931 \mathrm{MeV}$ )
1 $2.67 \mathrm{MeV}$
2 $26.7 \mathrm{MeV}$
3 $6.675 \mathrm{MeV}$
4 $13.35 \mathrm{MeV}$
Explanation:
C Given that, Mass of reaction $(\Delta \mathrm{m})=0.02866 \mathrm{u}$ $1 \mathrm{u}=931 \mathrm{MeV}$ The nuclear fusion reaction, ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$ Energy released $\left(\mathrm{E}_{\mathrm{T}}\right)=\Delta \mathrm{m} \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \mathrm{u} \times \frac{931 \frac{\mathrm{MeV}}{\mathrm{c}^{2}}}{\mathrm{u}} \times \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \times 931.5$ $\mathrm{E}_{\mathrm{T}}=26.697$ We know that atomic mass of helium is 4 . So, Total energy released per $\mathrm{u}$ is - $\mathrm{E}=\frac{\mathrm{E}_{\mathrm{T}}}{4}$ $\mathrm{E}=\frac{26.697}{4}$ $\mathrm{E}=6.675 \mathrm{MeV}$
148008
Calculate power output of ${ }_{92}^{235} U$ reactor, if it takes 30 days to use up $2 \mathrm{~kg}$ of fuel, and if each fission gives $185 \mathrm{MeV}$ of useable energy. A Avogadro's number $=6 \times 10^{23} / \mathrm{mol}$ ?
148011
A radioactive substance decays to $1 / 16^{\text {th }}$ of its initial activity in 40 days. The half-life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given that, Total time taken in decay $(\mathrm{t})=40$ days According to question - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{40}{\mathrm{~T}_{1 / 2}}}$ On comparing both the side, we get- $\frac{40}{\mathrm{~T}_{1 / 2}}=4$ $\mathrm{~T}_{1 / 2}=\frac{40}{4}$ $\mathrm{~T}_{1 / 2}=10 \text { days }$
AIIMS-2003
NUCLEAR PHYSICS
148015
The following equation represents induced transmutation ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\mathrm{X}$ In this equation, $X$ represents
1 one $\beta^{-}$particle
2 $\alpha$-particle
3 a positron
4 a neutron
Explanation:
D Given reaction, ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ Sum of atomic number in L.H.S $=4+2=6$ Sum of atomic number in R.H.S $=6+Z$ Then, $\text { L.H.S = R.H.S }$ $4+2=6+Z$ $Z=0$ Sum of mass number in L.H.S $=9+4=13$ Sum of mass number in R.H.S $=12+$ A Then, $13=12+A$ $A=13-12$ $A=1$ So, the emitted particle is neutron $\left({ }_{0} \mathrm{n}^{1}\right)$.
BCECE-2010
NUCLEAR PHYSICS
148016
A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866 \mathrm{u}$. The energy liberated per $u$ is (given $1 \mathrm{u} 931 \mathrm{MeV}$ )
1 $2.67 \mathrm{MeV}$
2 $26.7 \mathrm{MeV}$
3 $6.675 \mathrm{MeV}$
4 $13.35 \mathrm{MeV}$
Explanation:
C Given that, Mass of reaction $(\Delta \mathrm{m})=0.02866 \mathrm{u}$ $1 \mathrm{u}=931 \mathrm{MeV}$ The nuclear fusion reaction, ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$ Energy released $\left(\mathrm{E}_{\mathrm{T}}\right)=\Delta \mathrm{m} \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \mathrm{u} \times \frac{931 \frac{\mathrm{MeV}}{\mathrm{c}^{2}}}{\mathrm{u}} \times \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{T}}=0.02866 \times 931.5$ $\mathrm{E}_{\mathrm{T}}=26.697$ We know that atomic mass of helium is 4 . So, Total energy released per $\mathrm{u}$ is - $\mathrm{E}=\frac{\mathrm{E}_{\mathrm{T}}}{4}$ $\mathrm{E}=\frac{26.697}{4}$ $\mathrm{E}=6.675 \mathrm{MeV}$
