A We know that energy emitted photon $\mathrm{E}=\mathrm{hv}$ Photon momentum $(\mathrm{P})=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{hv}}{\mathrm{c}}$ According to momentum conservation value of recoiled nucleus should also be $\frac{\mathrm{hv}}{\mathrm{c}}$ but in opposite direction to make momentum zero. Now recoil energy $\left(E^{\prime}\right)=\frac{P^{2}}{2 M}$ $\mathrm{E}^{\prime}=\frac{\mathrm{P}^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\left(\frac{\mathrm{hv}}{\mathrm{c}}\right)^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{h}^{2} \mathrm{v}^{2}}{2 \mathrm{Mc}^{2}}$
AIPMT- 2011
NUCLEAR PHYSICS
147683
In the nuclear reaction ${ }_{7}^{14} \mathrm{~N}+\mathrm{X} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}$ the $X$ will be
1 ${ }_{-1}^{0} \mathrm{e}$
2 ${ }_{1}^{1} \mathrm{H}$
3 ${ }_{1}^{2} \mathrm{H}$
4 ${ }_{0}^{1} \mathrm{n}$
Explanation:
D The given nuclear reaction - ${ }_{7} \mathrm{~N}^{14}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{6} \mathrm{C}^{14}+{ }_{1} \mathrm{H}^{1}$ Applying conservation of mass number, $14+\mathrm{A}=14+1$ $\mathrm{~A}=1$ Applying conservation of atomic number, $7+Z=6+1$ $Z=0$ Then, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}={ }_{0} \mathrm{n}^{1}$
WB JEE 2011
NUCLEAR PHYSICS
147685
Two samples $X$ and $Y$ contain equal amount of radioactive substances. If $\frac{1}{16}$ th of the sample $X$ and $\frac{1}{256}$ th of the sample $Y$, remain after $8 \mathrm{~h}$, then the ratio of half periods of $\mathrm{X}$ and $\mathrm{Y}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $1 ; 16$
5 $4: 1$
Explanation:
A For sample $\mathrm{x}$, Time $(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $\mathrm{x}$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{16}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{x}} \times 8}=16$ $\lambda_{\mathrm{x}} \times 8=\ln 16$ $\lambda_{\mathrm{x}}=\frac{\ln 16}{8}$ $\lambda_{\mathrm{x}}=\frac{4 \ln 2}{8}=\frac{\ln 2}{2}$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\lambda_{\mathrm{x}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\ln 2} \times 2$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{x}}=2$ For sample $\mathrm{y}$, $\text { Time }(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $y$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{256}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{256}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{y}} \times 8}=256$ $\lambda_{\mathrm{y}} \times 8=\ln 256$ $\lambda_{\mathrm{y}}=\frac{\ln 256}{8}$ $\lambda_{\mathrm{y}}=\frac{8 \ln 2}{8}=\ln 2$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\lambda_{\mathrm{y}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\ln 2}$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{y}}=1$ On dividing equation (i) \& (ii), we get - $\frac{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}}{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}}=\frac{2}{1}=2: 1$
Kerala CEE - 2010
NUCLEAR PHYSICS
147686
Which one of the following statement is true, if half-life of a radioactive substance is 1 month?
1 $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months
2 $1 / 8^{\text {th }}$ part of the substance will remain undecayed at the end of 4 months
3 The substance will disintegrate completely in 4 months
4 $1 / 16^{\text {th }}$ part of the substance will remain undecayed at the end of 3 months
5 The substance will disintegrate completely in 2 months
Explanation:
A Given, $\mathrm{T}_{1 / 2}=1$ month $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left[\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right]$ For $\mathrm{t}=3$ months, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 1}=\frac{1}{8}$ Therefore, disintegrated part substance in 3 months- $=1-\frac{1}{8}$ $=\frac{7}{8}$ Hence, $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months.
A We know that energy emitted photon $\mathrm{E}=\mathrm{hv}$ Photon momentum $(\mathrm{P})=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{hv}}{\mathrm{c}}$ According to momentum conservation value of recoiled nucleus should also be $\frac{\mathrm{hv}}{\mathrm{c}}$ but in opposite direction to make momentum zero. Now recoil energy $\left(E^{\prime}\right)=\frac{P^{2}}{2 M}$ $\mathrm{E}^{\prime}=\frac{\mathrm{P}^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\left(\frac{\mathrm{hv}}{\mathrm{c}}\right)^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{h}^{2} \mathrm{v}^{2}}{2 \mathrm{Mc}^{2}}$
AIPMT- 2011
NUCLEAR PHYSICS
147683
In the nuclear reaction ${ }_{7}^{14} \mathrm{~N}+\mathrm{X} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}$ the $X$ will be
1 ${ }_{-1}^{0} \mathrm{e}$
2 ${ }_{1}^{1} \mathrm{H}$
3 ${ }_{1}^{2} \mathrm{H}$
4 ${ }_{0}^{1} \mathrm{n}$
Explanation:
D The given nuclear reaction - ${ }_{7} \mathrm{~N}^{14}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{6} \mathrm{C}^{14}+{ }_{1} \mathrm{H}^{1}$ Applying conservation of mass number, $14+\mathrm{A}=14+1$ $\mathrm{~A}=1$ Applying conservation of atomic number, $7+Z=6+1$ $Z=0$ Then, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}={ }_{0} \mathrm{n}^{1}$
WB JEE 2011
NUCLEAR PHYSICS
147685
Two samples $X$ and $Y$ contain equal amount of radioactive substances. If $\frac{1}{16}$ th of the sample $X$ and $\frac{1}{256}$ th of the sample $Y$, remain after $8 \mathrm{~h}$, then the ratio of half periods of $\mathrm{X}$ and $\mathrm{Y}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $1 ; 16$
5 $4: 1$
Explanation:
A For sample $\mathrm{x}$, Time $(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $\mathrm{x}$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{16}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{x}} \times 8}=16$ $\lambda_{\mathrm{x}} \times 8=\ln 16$ $\lambda_{\mathrm{x}}=\frac{\ln 16}{8}$ $\lambda_{\mathrm{x}}=\frac{4 \ln 2}{8}=\frac{\ln 2}{2}$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\lambda_{\mathrm{x}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\ln 2} \times 2$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{x}}=2$ For sample $\mathrm{y}$, $\text { Time }(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $y$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{256}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{256}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{y}} \times 8}=256$ $\lambda_{\mathrm{y}} \times 8=\ln 256$ $\lambda_{\mathrm{y}}=\frac{\ln 256}{8}$ $\lambda_{\mathrm{y}}=\frac{8 \ln 2}{8}=\ln 2$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\lambda_{\mathrm{y}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\ln 2}$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{y}}=1$ On dividing equation (i) \& (ii), we get - $\frac{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}}{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}}=\frac{2}{1}=2: 1$
Kerala CEE - 2010
NUCLEAR PHYSICS
147686
Which one of the following statement is true, if half-life of a radioactive substance is 1 month?
1 $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months
2 $1 / 8^{\text {th }}$ part of the substance will remain undecayed at the end of 4 months
3 The substance will disintegrate completely in 4 months
4 $1 / 16^{\text {th }}$ part of the substance will remain undecayed at the end of 3 months
5 The substance will disintegrate completely in 2 months
Explanation:
A Given, $\mathrm{T}_{1 / 2}=1$ month $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left[\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right]$ For $\mathrm{t}=3$ months, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 1}=\frac{1}{8}$ Therefore, disintegrated part substance in 3 months- $=1-\frac{1}{8}$ $=\frac{7}{8}$ Hence, $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months.
A We know that energy emitted photon $\mathrm{E}=\mathrm{hv}$ Photon momentum $(\mathrm{P})=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{hv}}{\mathrm{c}}$ According to momentum conservation value of recoiled nucleus should also be $\frac{\mathrm{hv}}{\mathrm{c}}$ but in opposite direction to make momentum zero. Now recoil energy $\left(E^{\prime}\right)=\frac{P^{2}}{2 M}$ $\mathrm{E}^{\prime}=\frac{\mathrm{P}^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\left(\frac{\mathrm{hv}}{\mathrm{c}}\right)^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{h}^{2} \mathrm{v}^{2}}{2 \mathrm{Mc}^{2}}$
AIPMT- 2011
NUCLEAR PHYSICS
147683
In the nuclear reaction ${ }_{7}^{14} \mathrm{~N}+\mathrm{X} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}$ the $X$ will be
1 ${ }_{-1}^{0} \mathrm{e}$
2 ${ }_{1}^{1} \mathrm{H}$
3 ${ }_{1}^{2} \mathrm{H}$
4 ${ }_{0}^{1} \mathrm{n}$
Explanation:
D The given nuclear reaction - ${ }_{7} \mathrm{~N}^{14}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{6} \mathrm{C}^{14}+{ }_{1} \mathrm{H}^{1}$ Applying conservation of mass number, $14+\mathrm{A}=14+1$ $\mathrm{~A}=1$ Applying conservation of atomic number, $7+Z=6+1$ $Z=0$ Then, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}={ }_{0} \mathrm{n}^{1}$
WB JEE 2011
NUCLEAR PHYSICS
147685
Two samples $X$ and $Y$ contain equal amount of radioactive substances. If $\frac{1}{16}$ th of the sample $X$ and $\frac{1}{256}$ th of the sample $Y$, remain after $8 \mathrm{~h}$, then the ratio of half periods of $\mathrm{X}$ and $\mathrm{Y}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $1 ; 16$
5 $4: 1$
Explanation:
A For sample $\mathrm{x}$, Time $(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $\mathrm{x}$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{16}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{x}} \times 8}=16$ $\lambda_{\mathrm{x}} \times 8=\ln 16$ $\lambda_{\mathrm{x}}=\frac{\ln 16}{8}$ $\lambda_{\mathrm{x}}=\frac{4 \ln 2}{8}=\frac{\ln 2}{2}$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\lambda_{\mathrm{x}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\ln 2} \times 2$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{x}}=2$ For sample $\mathrm{y}$, $\text { Time }(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $y$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{256}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{256}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{y}} \times 8}=256$ $\lambda_{\mathrm{y}} \times 8=\ln 256$ $\lambda_{\mathrm{y}}=\frac{\ln 256}{8}$ $\lambda_{\mathrm{y}}=\frac{8 \ln 2}{8}=\ln 2$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\lambda_{\mathrm{y}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\ln 2}$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{y}}=1$ On dividing equation (i) \& (ii), we get - $\frac{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}}{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}}=\frac{2}{1}=2: 1$
Kerala CEE - 2010
NUCLEAR PHYSICS
147686
Which one of the following statement is true, if half-life of a radioactive substance is 1 month?
1 $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months
2 $1 / 8^{\text {th }}$ part of the substance will remain undecayed at the end of 4 months
3 The substance will disintegrate completely in 4 months
4 $1 / 16^{\text {th }}$ part of the substance will remain undecayed at the end of 3 months
5 The substance will disintegrate completely in 2 months
Explanation:
A Given, $\mathrm{T}_{1 / 2}=1$ month $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left[\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right]$ For $\mathrm{t}=3$ months, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 1}=\frac{1}{8}$ Therefore, disintegrated part substance in 3 months- $=1-\frac{1}{8}$ $=\frac{7}{8}$ Hence, $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months.
A We know that energy emitted photon $\mathrm{E}=\mathrm{hv}$ Photon momentum $(\mathrm{P})=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{hv}}{\mathrm{c}}$ According to momentum conservation value of recoiled nucleus should also be $\frac{\mathrm{hv}}{\mathrm{c}}$ but in opposite direction to make momentum zero. Now recoil energy $\left(E^{\prime}\right)=\frac{P^{2}}{2 M}$ $\mathrm{E}^{\prime}=\frac{\mathrm{P}^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\left(\frac{\mathrm{hv}}{\mathrm{c}}\right)^{2}}{2 \mathrm{M}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{h}^{2} \mathrm{v}^{2}}{2 \mathrm{Mc}^{2}}$
AIPMT- 2011
NUCLEAR PHYSICS
147683
In the nuclear reaction ${ }_{7}^{14} \mathrm{~N}+\mathrm{X} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}$ the $X$ will be
1 ${ }_{-1}^{0} \mathrm{e}$
2 ${ }_{1}^{1} \mathrm{H}$
3 ${ }_{1}^{2} \mathrm{H}$
4 ${ }_{0}^{1} \mathrm{n}$
Explanation:
D The given nuclear reaction - ${ }_{7} \mathrm{~N}^{14}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{6} \mathrm{C}^{14}+{ }_{1} \mathrm{H}^{1}$ Applying conservation of mass number, $14+\mathrm{A}=14+1$ $\mathrm{~A}=1$ Applying conservation of atomic number, $7+Z=6+1$ $Z=0$ Then, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}={ }_{0} \mathrm{n}^{1}$
WB JEE 2011
NUCLEAR PHYSICS
147685
Two samples $X$ and $Y$ contain equal amount of radioactive substances. If $\frac{1}{16}$ th of the sample $X$ and $\frac{1}{256}$ th of the sample $Y$, remain after $8 \mathrm{~h}$, then the ratio of half periods of $\mathrm{X}$ and $\mathrm{Y}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $1 ; 16$
5 $4: 1$
Explanation:
A For sample $\mathrm{x}$, Time $(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $\mathrm{x}$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{16}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{x}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{x}} \times 8}=16$ $\lambda_{\mathrm{x}} \times 8=\ln 16$ $\lambda_{\mathrm{x}}=\frac{\ln 16}{8}$ $\lambda_{\mathrm{x}}=\frac{4 \ln 2}{8}=\frac{\ln 2}{2}$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\lambda_{\mathrm{x}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}=\frac{\ln 2}{\ln 2} \times 2$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{x}}=2$ For sample $\mathrm{y}$, $\text { Time }(\mathrm{t})=8 \mathrm{~h}$ Remaining amount of sample $y$, $\mathrm{N}=\frac{\mathrm{N}_{0}}{256}$ Then, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{256}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{y}} \times 8}$ $\mathrm{e}^{\lambda_{\mathrm{y}} \times 8}=256$ $\lambda_{\mathrm{y}} \times 8=\ln 256$ $\lambda_{\mathrm{y}}=\frac{\ln 256}{8}$ $\lambda_{\mathrm{y}}=\frac{8 \ln 2}{8}=\ln 2$ Half life $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\lambda_{\mathrm{y}}}$ $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}=\frac{\ln 2}{\ln 2}$ $\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{y}}=1$ On dividing equation (i) \& (ii), we get - $\frac{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{x}}}{\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{y}}}=\frac{2}{1}=2: 1$
Kerala CEE - 2010
NUCLEAR PHYSICS
147686
Which one of the following statement is true, if half-life of a radioactive substance is 1 month?
1 $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months
2 $1 / 8^{\text {th }}$ part of the substance will remain undecayed at the end of 4 months
3 The substance will disintegrate completely in 4 months
4 $1 / 16^{\text {th }}$ part of the substance will remain undecayed at the end of 3 months
5 The substance will disintegrate completely in 2 months
Explanation:
A Given, $\mathrm{T}_{1 / 2}=1$ month $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left[\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right]$ For $\mathrm{t}=3$ months, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 1}=\frac{1}{8}$ Therefore, disintegrated part substance in 3 months- $=1-\frac{1}{8}$ $=\frac{7}{8}$ Hence, $7 / 8^{\text {th }}$ part of the substance will disintegrate in 3 months.