363624
A nucleus with mass number 220 initially at rest emits an \(\alpha \)-particle. If the \(Q\) value of the reaction is \(5.5\,MeV\), calculate the kinetic energy of the \(\alpha \)-particle:
1 \(4.4\,MeV\)
2 \(5.4\,MeV\)
3 \(5.6\,MeV\)
4 \(4.5\,MeV\)
Explanation:
Linear momentum is conserved. \(\therefore {p_1} = {p_2}\) But \(p = \sqrt {2mK} \) where \(K = \) kinetic energy \(\therefore \sqrt {2(216m){K_1}} = \sqrt {2(4m){K_2}} \) \(216{K_1} = 4{K_2} \Rightarrow {K_2} = 54{K_1}\) (1) Given \({K_1} + {K_2} = 5.5MeV\) (2) From (1) and (2) or \({K_1} = \frac{1}{{10}}MeV\) and \({K_2} = 5.4MeV\) \(\therefore \) Kinetic energy of \(\alpha \) particle \( = 5.4\,MeV\).
PHXII13:NUCLEI
363625
If total energy of electron is \(3.555\,MeV\) then its rest mass energy is
363627
What is the \(Q\)-value of the reaction \(p+{ }^{7} \mathrm{Li} \rightarrow{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}\) The atomic masses of \(^1H{,^4}He\) and \(^7Li\) are \(1.007825\,u,4.002603\,u\) and \(7.016004\,u\) respectively
1 \(17.35\,MeV\)
2 \(18.06\,MeV\)
3 \(177.35\,MeV\)
4 \(170.35\,MeV\)
Explanation:
The total mass of the initial particles \(m_{i}=1.007825+7.016004=8.023829 u\) and the total mass of the initial and final mass of particles. \(\begin{aligned}m_{f} & =2 \times 4.002603=8.005206 \\\Delta m & =m_{i}-m_{f} \\& =8.023829-8.005206=0.018623 u\end{aligned}\) The \(Q\) - value is given by \(Q = (\Delta m){c^2} = 0.018623 \times 931.5\) \( = 17.35\,{\rm{MeV}}\).
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PHXII13:NUCLEI
363624
A nucleus with mass number 220 initially at rest emits an \(\alpha \)-particle. If the \(Q\) value of the reaction is \(5.5\,MeV\), calculate the kinetic energy of the \(\alpha \)-particle:
1 \(4.4\,MeV\)
2 \(5.4\,MeV\)
3 \(5.6\,MeV\)
4 \(4.5\,MeV\)
Explanation:
Linear momentum is conserved. \(\therefore {p_1} = {p_2}\) But \(p = \sqrt {2mK} \) where \(K = \) kinetic energy \(\therefore \sqrt {2(216m){K_1}} = \sqrt {2(4m){K_2}} \) \(216{K_1} = 4{K_2} \Rightarrow {K_2} = 54{K_1}\) (1) Given \({K_1} + {K_2} = 5.5MeV\) (2) From (1) and (2) or \({K_1} = \frac{1}{{10}}MeV\) and \({K_2} = 5.4MeV\) \(\therefore \) Kinetic energy of \(\alpha \) particle \( = 5.4\,MeV\).
PHXII13:NUCLEI
363625
If total energy of electron is \(3.555\,MeV\) then its rest mass energy is
363627
What is the \(Q\)-value of the reaction \(p+{ }^{7} \mathrm{Li} \rightarrow{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}\) The atomic masses of \(^1H{,^4}He\) and \(^7Li\) are \(1.007825\,u,4.002603\,u\) and \(7.016004\,u\) respectively
1 \(17.35\,MeV\)
2 \(18.06\,MeV\)
3 \(177.35\,MeV\)
4 \(170.35\,MeV\)
Explanation:
The total mass of the initial particles \(m_{i}=1.007825+7.016004=8.023829 u\) and the total mass of the initial and final mass of particles. \(\begin{aligned}m_{f} & =2 \times 4.002603=8.005206 \\\Delta m & =m_{i}-m_{f} \\& =8.023829-8.005206=0.018623 u\end{aligned}\) The \(Q\) - value is given by \(Q = (\Delta m){c^2} = 0.018623 \times 931.5\) \( = 17.35\,{\rm{MeV}}\).
363624
A nucleus with mass number 220 initially at rest emits an \(\alpha \)-particle. If the \(Q\) value of the reaction is \(5.5\,MeV\), calculate the kinetic energy of the \(\alpha \)-particle:
1 \(4.4\,MeV\)
2 \(5.4\,MeV\)
3 \(5.6\,MeV\)
4 \(4.5\,MeV\)
Explanation:
Linear momentum is conserved. \(\therefore {p_1} = {p_2}\) But \(p = \sqrt {2mK} \) where \(K = \) kinetic energy \(\therefore \sqrt {2(216m){K_1}} = \sqrt {2(4m){K_2}} \) \(216{K_1} = 4{K_2} \Rightarrow {K_2} = 54{K_1}\) (1) Given \({K_1} + {K_2} = 5.5MeV\) (2) From (1) and (2) or \({K_1} = \frac{1}{{10}}MeV\) and \({K_2} = 5.4MeV\) \(\therefore \) Kinetic energy of \(\alpha \) particle \( = 5.4\,MeV\).
PHXII13:NUCLEI
363625
If total energy of electron is \(3.555\,MeV\) then its rest mass energy is
363627
What is the \(Q\)-value of the reaction \(p+{ }^{7} \mathrm{Li} \rightarrow{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}\) The atomic masses of \(^1H{,^4}He\) and \(^7Li\) are \(1.007825\,u,4.002603\,u\) and \(7.016004\,u\) respectively
1 \(17.35\,MeV\)
2 \(18.06\,MeV\)
3 \(177.35\,MeV\)
4 \(170.35\,MeV\)
Explanation:
The total mass of the initial particles \(m_{i}=1.007825+7.016004=8.023829 u\) and the total mass of the initial and final mass of particles. \(\begin{aligned}m_{f} & =2 \times 4.002603=8.005206 \\\Delta m & =m_{i}-m_{f} \\& =8.023829-8.005206=0.018623 u\end{aligned}\) The \(Q\) - value is given by \(Q = (\Delta m){c^2} = 0.018623 \times 931.5\) \( = 17.35\,{\rm{MeV}}\).
363624
A nucleus with mass number 220 initially at rest emits an \(\alpha \)-particle. If the \(Q\) value of the reaction is \(5.5\,MeV\), calculate the kinetic energy of the \(\alpha \)-particle:
1 \(4.4\,MeV\)
2 \(5.4\,MeV\)
3 \(5.6\,MeV\)
4 \(4.5\,MeV\)
Explanation:
Linear momentum is conserved. \(\therefore {p_1} = {p_2}\) But \(p = \sqrt {2mK} \) where \(K = \) kinetic energy \(\therefore \sqrt {2(216m){K_1}} = \sqrt {2(4m){K_2}} \) \(216{K_1} = 4{K_2} \Rightarrow {K_2} = 54{K_1}\) (1) Given \({K_1} + {K_2} = 5.5MeV\) (2) From (1) and (2) or \({K_1} = \frac{1}{{10}}MeV\) and \({K_2} = 5.4MeV\) \(\therefore \) Kinetic energy of \(\alpha \) particle \( = 5.4\,MeV\).
PHXII13:NUCLEI
363625
If total energy of electron is \(3.555\,MeV\) then its rest mass energy is
363627
What is the \(Q\)-value of the reaction \(p+{ }^{7} \mathrm{Li} \rightarrow{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}\) The atomic masses of \(^1H{,^4}He\) and \(^7Li\) are \(1.007825\,u,4.002603\,u\) and \(7.016004\,u\) respectively
1 \(17.35\,MeV\)
2 \(18.06\,MeV\)
3 \(177.35\,MeV\)
4 \(170.35\,MeV\)
Explanation:
The total mass of the initial particles \(m_{i}=1.007825+7.016004=8.023829 u\) and the total mass of the initial and final mass of particles. \(\begin{aligned}m_{f} & =2 \times 4.002603=8.005206 \\\Delta m & =m_{i}-m_{f} \\& =8.023829-8.005206=0.018623 u\end{aligned}\) The \(Q\) - value is given by \(Q = (\Delta m){c^2} = 0.018623 \times 931.5\) \( = 17.35\,{\rm{MeV}}\).