147639
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number to the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \beta, 7 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D ${ }_{90} \mathrm{Th}^{228} \rightarrow{ }_{83} \mathrm{Bi}^{212}$ Mass number is reduced by $=228-212$ $=16$ Number of alpha particle emitted are $\left(\mathrm{n}_{\alpha}\right)=\frac{16}{4}=4$ Number of $\beta$-particle emitted are $\left(\mathrm{n}_{\beta}\right)$ - $=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ 4- $\alpha$ particle and 1- $\beta$ particle are emitted.
Manipal UGET-2014
NUCLEAR PHYSICS
147640
The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$. The fraction that decays is 10 days will be
1 $77 / 80$
2 $71 / 80$
3 $31 / 32$
4 $15 / 16$
Explanation:
C Number of radioactive element that decays $=\frac{7}{8}$ Number of radioactive element that left decays $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=$ $=1-\frac{7}{8}=\frac{1}{8}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{t}=\mathrm{T}_{1 / 2} \frac{\log \mathrm{e}\left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)}{\log 2}$ $\mathrm{t}=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln 2}$ $6=\frac{\mathrm{T}_{1 / 2} \ln (8 / 1)}{\ln 2}$ $10=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln (2)}$ On dividing equation (i) and (ii), we get- $\frac{6}{10}=\frac{\ln (8)}{\ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\frac{5}{3} \ln (8)$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (8)^{5 / 3}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (32)$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=32$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{32}$ So, the fraction that decays is 10 days will be- $=1-\frac{1}{32}=\frac{31}{32}$
Manipal UGET-2014
NUCLEAR PHYSICS
147641
A nucleus $X$ initially at rest, undergoes alpha decay according to the equation ${ }_{92} \mathbf{X}^{\mathrm{A}} \rightarrow_{\mathrm{Z}} \mathbf{Y}^{228}+\alpha$ Then, the value of $A$ and $Z$ are
1 94,230
2 232,90
3 190,32
4 230,94
Explanation:
B The equation is given by, ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+\alpha$ We can write above equation as- ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+{ }_{2} \mathrm{He}^{4} \therefore \alpha={ }_{2} \mathrm{He}^{4}$ Let atomic number and mass number is $\mathrm{Z}$ and $\mathrm{A}$. So, $\mathrm{A}=228+4$ $\mathrm{A}=232$ And, $\quad 92=Z+2$ $Z=90$ Hence, the value of $A$ and $Z$ are 232 and 90 .
Manipal UGET-2012
NUCLEAR PHYSICS
147642
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is
1 $80 \mathrm{~min}$
2 $120 \mathrm{~min}$
3 $20 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
C Given, \(\mathrm{N}_{\mathrm{o}}=240\) per minute \(\mathrm{N}=30\) per minute Time \((\mathrm{t})=1 \mathrm{hr}=60\) minute The equation for initial and final count rate is- \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Where, \(\mathrm{N}=\) final count rate \(\mathrm{N}_{\mathrm{o}}=\) initial count rate \(\mathrm{n}=\) number of half lives Then, \(30=240\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{30}{240}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{60}{t_{1 / 2}}=3\) \(t_{1 / 2}=20 \mathrm{~min}\)
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NUCLEAR PHYSICS
147639
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number to the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \beta, 7 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D ${ }_{90} \mathrm{Th}^{228} \rightarrow{ }_{83} \mathrm{Bi}^{212}$ Mass number is reduced by $=228-212$ $=16$ Number of alpha particle emitted are $\left(\mathrm{n}_{\alpha}\right)=\frac{16}{4}=4$ Number of $\beta$-particle emitted are $\left(\mathrm{n}_{\beta}\right)$ - $=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ 4- $\alpha$ particle and 1- $\beta$ particle are emitted.
Manipal UGET-2014
NUCLEAR PHYSICS
147640
The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$. The fraction that decays is 10 days will be
1 $77 / 80$
2 $71 / 80$
3 $31 / 32$
4 $15 / 16$
Explanation:
C Number of radioactive element that decays $=\frac{7}{8}$ Number of radioactive element that left decays $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=$ $=1-\frac{7}{8}=\frac{1}{8}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{t}=\mathrm{T}_{1 / 2} \frac{\log \mathrm{e}\left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)}{\log 2}$ $\mathrm{t}=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln 2}$ $6=\frac{\mathrm{T}_{1 / 2} \ln (8 / 1)}{\ln 2}$ $10=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln (2)}$ On dividing equation (i) and (ii), we get- $\frac{6}{10}=\frac{\ln (8)}{\ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\frac{5}{3} \ln (8)$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (8)^{5 / 3}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (32)$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=32$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{32}$ So, the fraction that decays is 10 days will be- $=1-\frac{1}{32}=\frac{31}{32}$
Manipal UGET-2014
NUCLEAR PHYSICS
147641
A nucleus $X$ initially at rest, undergoes alpha decay according to the equation ${ }_{92} \mathbf{X}^{\mathrm{A}} \rightarrow_{\mathrm{Z}} \mathbf{Y}^{228}+\alpha$ Then, the value of $A$ and $Z$ are
1 94,230
2 232,90
3 190,32
4 230,94
Explanation:
B The equation is given by, ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+\alpha$ We can write above equation as- ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+{ }_{2} \mathrm{He}^{4} \therefore \alpha={ }_{2} \mathrm{He}^{4}$ Let atomic number and mass number is $\mathrm{Z}$ and $\mathrm{A}$. So, $\mathrm{A}=228+4$ $\mathrm{A}=232$ And, $\quad 92=Z+2$ $Z=90$ Hence, the value of $A$ and $Z$ are 232 and 90 .
Manipal UGET-2012
NUCLEAR PHYSICS
147642
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is
1 $80 \mathrm{~min}$
2 $120 \mathrm{~min}$
3 $20 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
C Given, \(\mathrm{N}_{\mathrm{o}}=240\) per minute \(\mathrm{N}=30\) per minute Time \((\mathrm{t})=1 \mathrm{hr}=60\) minute The equation for initial and final count rate is- \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Where, \(\mathrm{N}=\) final count rate \(\mathrm{N}_{\mathrm{o}}=\) initial count rate \(\mathrm{n}=\) number of half lives Then, \(30=240\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{30}{240}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{60}{t_{1 / 2}}=3\) \(t_{1 / 2}=20 \mathrm{~min}\)
147639
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number to the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \beta, 7 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D ${ }_{90} \mathrm{Th}^{228} \rightarrow{ }_{83} \mathrm{Bi}^{212}$ Mass number is reduced by $=228-212$ $=16$ Number of alpha particle emitted are $\left(\mathrm{n}_{\alpha}\right)=\frac{16}{4}=4$ Number of $\beta$-particle emitted are $\left(\mathrm{n}_{\beta}\right)$ - $=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ 4- $\alpha$ particle and 1- $\beta$ particle are emitted.
Manipal UGET-2014
NUCLEAR PHYSICS
147640
The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$. The fraction that decays is 10 days will be
1 $77 / 80$
2 $71 / 80$
3 $31 / 32$
4 $15 / 16$
Explanation:
C Number of radioactive element that decays $=\frac{7}{8}$ Number of radioactive element that left decays $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=$ $=1-\frac{7}{8}=\frac{1}{8}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{t}=\mathrm{T}_{1 / 2} \frac{\log \mathrm{e}\left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)}{\log 2}$ $\mathrm{t}=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln 2}$ $6=\frac{\mathrm{T}_{1 / 2} \ln (8 / 1)}{\ln 2}$ $10=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln (2)}$ On dividing equation (i) and (ii), we get- $\frac{6}{10}=\frac{\ln (8)}{\ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\frac{5}{3} \ln (8)$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (8)^{5 / 3}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (32)$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=32$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{32}$ So, the fraction that decays is 10 days will be- $=1-\frac{1}{32}=\frac{31}{32}$
Manipal UGET-2014
NUCLEAR PHYSICS
147641
A nucleus $X$ initially at rest, undergoes alpha decay according to the equation ${ }_{92} \mathbf{X}^{\mathrm{A}} \rightarrow_{\mathrm{Z}} \mathbf{Y}^{228}+\alpha$ Then, the value of $A$ and $Z$ are
1 94,230
2 232,90
3 190,32
4 230,94
Explanation:
B The equation is given by, ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+\alpha$ We can write above equation as- ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+{ }_{2} \mathrm{He}^{4} \therefore \alpha={ }_{2} \mathrm{He}^{4}$ Let atomic number and mass number is $\mathrm{Z}$ and $\mathrm{A}$. So, $\mathrm{A}=228+4$ $\mathrm{A}=232$ And, $\quad 92=Z+2$ $Z=90$ Hence, the value of $A$ and $Z$ are 232 and 90 .
Manipal UGET-2012
NUCLEAR PHYSICS
147642
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is
1 $80 \mathrm{~min}$
2 $120 \mathrm{~min}$
3 $20 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
C Given, \(\mathrm{N}_{\mathrm{o}}=240\) per minute \(\mathrm{N}=30\) per minute Time \((\mathrm{t})=1 \mathrm{hr}=60\) minute The equation for initial and final count rate is- \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Where, \(\mathrm{N}=\) final count rate \(\mathrm{N}_{\mathrm{o}}=\) initial count rate \(\mathrm{n}=\) number of half lives Then, \(30=240\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{30}{240}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{60}{t_{1 / 2}}=3\) \(t_{1 / 2}=20 \mathrm{~min}\)
147639
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number to the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \beta, 7 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D ${ }_{90} \mathrm{Th}^{228} \rightarrow{ }_{83} \mathrm{Bi}^{212}$ Mass number is reduced by $=228-212$ $=16$ Number of alpha particle emitted are $\left(\mathrm{n}_{\alpha}\right)=\frac{16}{4}=4$ Number of $\beta$-particle emitted are $\left(\mathrm{n}_{\beta}\right)$ - $=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ 4- $\alpha$ particle and 1- $\beta$ particle are emitted.
Manipal UGET-2014
NUCLEAR PHYSICS
147640
The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$. The fraction that decays is 10 days will be
1 $77 / 80$
2 $71 / 80$
3 $31 / 32$
4 $15 / 16$
Explanation:
C Number of radioactive element that decays $=\frac{7}{8}$ Number of radioactive element that left decays $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=$ $=1-\frac{7}{8}=\frac{1}{8}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{t}=\mathrm{T}_{1 / 2} \frac{\log \mathrm{e}\left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)}{\log 2}$ $\mathrm{t}=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln 2}$ $6=\frac{\mathrm{T}_{1 / 2} \ln (8 / 1)}{\ln 2}$ $10=\frac{\mathrm{T}_{1 / 2} \ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}{\ln (2)}$ On dividing equation (i) and (ii), we get- $\frac{6}{10}=\frac{\ln (8)}{\ln \left(\mathrm{N}_{0} / \mathrm{N}\right)}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\frac{5}{3} \ln (8)$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (8)^{5 / 3}$ $\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)=\ln (32)$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=32$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{32}$ So, the fraction that decays is 10 days will be- $=1-\frac{1}{32}=\frac{31}{32}$
Manipal UGET-2014
NUCLEAR PHYSICS
147641
A nucleus $X$ initially at rest, undergoes alpha decay according to the equation ${ }_{92} \mathbf{X}^{\mathrm{A}} \rightarrow_{\mathrm{Z}} \mathbf{Y}^{228}+\alpha$ Then, the value of $A$ and $Z$ are
1 94,230
2 232,90
3 190,32
4 230,94
Explanation:
B The equation is given by, ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+\alpha$ We can write above equation as- ${ }_{92} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}} \mathrm{Y}^{228}+{ }_{2} \mathrm{He}^{4} \therefore \alpha={ }_{2} \mathrm{He}^{4}$ Let atomic number and mass number is $\mathrm{Z}$ and $\mathrm{A}$. So, $\mathrm{A}=228+4$ $\mathrm{A}=232$ And, $\quad 92=Z+2$ $Z=90$ Hence, the value of $A$ and $Z$ are 232 and 90 .
Manipal UGET-2012
NUCLEAR PHYSICS
147642
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is
1 $80 \mathrm{~min}$
2 $120 \mathrm{~min}$
3 $20 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
C Given, \(\mathrm{N}_{\mathrm{o}}=240\) per minute \(\mathrm{N}=30\) per minute Time \((\mathrm{t})=1 \mathrm{hr}=60\) minute The equation for initial and final count rate is- \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Where, \(\mathrm{N}=\) final count rate \(\mathrm{N}_{\mathrm{o}}=\) initial count rate \(\mathrm{n}=\) number of half lives Then, \(30=240\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{30}{240}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\frac{60}{t_{1 / 2}}}\) \(\frac{60}{t_{1 / 2}}=3\) \(t_{1 / 2}=20 \mathrm{~min}\)
