364025
In a sample of rock there are two radioactive elements \(A\) and \(B\) having initial concentration ratio 1:4. If \({\lambda _A}\) and \({\lambda _B}\) are rate constants then the time at which the concentration of both \(A\) and \(B\) are equal.
The concentrations of \(A\) and \(B\) at time \(t\) are \({N_A} = {N_{{0_A}}}{e^{ - {\lambda _A}t}}\) & \({N_B} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) Given that \(\frac{{{N_{{o_A}}}}}{{{N_{{0_B}}}}} = \frac{1}{4}\) \({N_{{0_A}}}{e^{ - {\lambda _A}t}} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) \(\frac{1}{4} = {e^{({\lambda _A} - {\lambda _B})t}}\) \(\ln \left( {\frac{1}{4}} \right) = ({\lambda _A} - {\lambda _B})t \Rightarrow t = \frac{1}{{{\lambda _A} - {\lambda _B}}}\ln \left( {\frac{1}{4}} \right)\)
PHXII13:NUCLEI
364026
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
364027
The half life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
1 40 minutes
2 10 minutes
3 30 minutes
4 25 minutes
Explanation:
After 50% decay, 50% \( = 1/2\) of the original material remains, i.e., the original material has undergone one half life \( = 20\) minutes. After 87.5% decay, 12.5%\( = 1/8\) of the original material remains. Since \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\), the original material has undergone 3 consecutive half lives \( = 20 \times 3 = 60\) minutes Therefore, time difference between 50% decay and 87.5% decay \( = 60 - 20 = 40\) minutres.
KCET - 2015
PHXII13:NUCLEI
364028
The activity of a radioactive element decreases to one third of the original activity \({A_{0}}\) in a period of nine years. After a further lapse of 9 years, if activity is found to be \({\dfrac{A_{0}}{N}}\). Find the value of \({N}\).
364025
In a sample of rock there are two radioactive elements \(A\) and \(B\) having initial concentration ratio 1:4. If \({\lambda _A}\) and \({\lambda _B}\) are rate constants then the time at which the concentration of both \(A\) and \(B\) are equal.
The concentrations of \(A\) and \(B\) at time \(t\) are \({N_A} = {N_{{0_A}}}{e^{ - {\lambda _A}t}}\) & \({N_B} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) Given that \(\frac{{{N_{{o_A}}}}}{{{N_{{0_B}}}}} = \frac{1}{4}\) \({N_{{0_A}}}{e^{ - {\lambda _A}t}} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) \(\frac{1}{4} = {e^{({\lambda _A} - {\lambda _B})t}}\) \(\ln \left( {\frac{1}{4}} \right) = ({\lambda _A} - {\lambda _B})t \Rightarrow t = \frac{1}{{{\lambda _A} - {\lambda _B}}}\ln \left( {\frac{1}{4}} \right)\)
PHXII13:NUCLEI
364026
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
364027
The half life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
1 40 minutes
2 10 minutes
3 30 minutes
4 25 minutes
Explanation:
After 50% decay, 50% \( = 1/2\) of the original material remains, i.e., the original material has undergone one half life \( = 20\) minutes. After 87.5% decay, 12.5%\( = 1/8\) of the original material remains. Since \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\), the original material has undergone 3 consecutive half lives \( = 20 \times 3 = 60\) minutes Therefore, time difference between 50% decay and 87.5% decay \( = 60 - 20 = 40\) minutres.
KCET - 2015
PHXII13:NUCLEI
364028
The activity of a radioactive element decreases to one third of the original activity \({A_{0}}\) in a period of nine years. After a further lapse of 9 years, if activity is found to be \({\dfrac{A_{0}}{N}}\). Find the value of \({N}\).
364025
In a sample of rock there are two radioactive elements \(A\) and \(B\) having initial concentration ratio 1:4. If \({\lambda _A}\) and \({\lambda _B}\) are rate constants then the time at which the concentration of both \(A\) and \(B\) are equal.
The concentrations of \(A\) and \(B\) at time \(t\) are \({N_A} = {N_{{0_A}}}{e^{ - {\lambda _A}t}}\) & \({N_B} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) Given that \(\frac{{{N_{{o_A}}}}}{{{N_{{0_B}}}}} = \frac{1}{4}\) \({N_{{0_A}}}{e^{ - {\lambda _A}t}} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) \(\frac{1}{4} = {e^{({\lambda _A} - {\lambda _B})t}}\) \(\ln \left( {\frac{1}{4}} \right) = ({\lambda _A} - {\lambda _B})t \Rightarrow t = \frac{1}{{{\lambda _A} - {\lambda _B}}}\ln \left( {\frac{1}{4}} \right)\)
PHXII13:NUCLEI
364026
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
364027
The half life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
1 40 minutes
2 10 minutes
3 30 minutes
4 25 minutes
Explanation:
After 50% decay, 50% \( = 1/2\) of the original material remains, i.e., the original material has undergone one half life \( = 20\) minutes. After 87.5% decay, 12.5%\( = 1/8\) of the original material remains. Since \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\), the original material has undergone 3 consecutive half lives \( = 20 \times 3 = 60\) minutes Therefore, time difference between 50% decay and 87.5% decay \( = 60 - 20 = 40\) minutres.
KCET - 2015
PHXII13:NUCLEI
364028
The activity of a radioactive element decreases to one third of the original activity \({A_{0}}\) in a period of nine years. After a further lapse of 9 years, if activity is found to be \({\dfrac{A_{0}}{N}}\). Find the value of \({N}\).
364025
In a sample of rock there are two radioactive elements \(A\) and \(B\) having initial concentration ratio 1:4. If \({\lambda _A}\) and \({\lambda _B}\) are rate constants then the time at which the concentration of both \(A\) and \(B\) are equal.
The concentrations of \(A\) and \(B\) at time \(t\) are \({N_A} = {N_{{0_A}}}{e^{ - {\lambda _A}t}}\) & \({N_B} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) Given that \(\frac{{{N_{{o_A}}}}}{{{N_{{0_B}}}}} = \frac{1}{4}\) \({N_{{0_A}}}{e^{ - {\lambda _A}t}} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) \(\frac{1}{4} = {e^{({\lambda _A} - {\lambda _B})t}}\) \(\ln \left( {\frac{1}{4}} \right) = ({\lambda _A} - {\lambda _B})t \Rightarrow t = \frac{1}{{{\lambda _A} - {\lambda _B}}}\ln \left( {\frac{1}{4}} \right)\)
PHXII13:NUCLEI
364026
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
364027
The half life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
1 40 minutes
2 10 minutes
3 30 minutes
4 25 minutes
Explanation:
After 50% decay, 50% \( = 1/2\) of the original material remains, i.e., the original material has undergone one half life \( = 20\) minutes. After 87.5% decay, 12.5%\( = 1/8\) of the original material remains. Since \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\), the original material has undergone 3 consecutive half lives \( = 20 \times 3 = 60\) minutes Therefore, time difference between 50% decay and 87.5% decay \( = 60 - 20 = 40\) minutres.
KCET - 2015
PHXII13:NUCLEI
364028
The activity of a radioactive element decreases to one third of the original activity \({A_{0}}\) in a period of nine years. After a further lapse of 9 years, if activity is found to be \({\dfrac{A_{0}}{N}}\). Find the value of \({N}\).
364025
In a sample of rock there are two radioactive elements \(A\) and \(B\) having initial concentration ratio 1:4. If \({\lambda _A}\) and \({\lambda _B}\) are rate constants then the time at which the concentration of both \(A\) and \(B\) are equal.
The concentrations of \(A\) and \(B\) at time \(t\) are \({N_A} = {N_{{0_A}}}{e^{ - {\lambda _A}t}}\) & \({N_B} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) Given that \(\frac{{{N_{{o_A}}}}}{{{N_{{0_B}}}}} = \frac{1}{4}\) \({N_{{0_A}}}{e^{ - {\lambda _A}t}} = {N_{{o_B}}}{e^{ - {\lambda _B}t}}\) \(\frac{1}{4} = {e^{({\lambda _A} - {\lambda _B})t}}\) \(\ln \left( {\frac{1}{4}} \right) = ({\lambda _A} - {\lambda _B})t \Rightarrow t = \frac{1}{{{\lambda _A} - {\lambda _B}}}\ln \left( {\frac{1}{4}} \right)\)
PHXII13:NUCLEI
364026
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
364027
The half life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
1 40 minutes
2 10 minutes
3 30 minutes
4 25 minutes
Explanation:
After 50% decay, 50% \( = 1/2\) of the original material remains, i.e., the original material has undergone one half life \( = 20\) minutes. After 87.5% decay, 12.5%\( = 1/8\) of the original material remains. Since \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\), the original material has undergone 3 consecutive half lives \( = 20 \times 3 = 60\) minutes Therefore, time difference between 50% decay and 87.5% decay \( = 60 - 20 = 40\) minutres.
KCET - 2015
PHXII13:NUCLEI
364028
The activity of a radioactive element decreases to one third of the original activity \({A_{0}}\) in a period of nine years. After a further lapse of 9 years, if activity is found to be \({\dfrac{A_{0}}{N}}\). Find the value of \({N}\).
