If the binding energy per nucleon in 73Li and He nuclei are 5.60MeV and 7.06MeV respectively, then in the reaction p+37Li→24 He, binding energy is
39.2 MeV
17.28 MeV
28.24 MeV
1.46 MeV
A
28.24 MeV
B
1.46 MeV
C
39.2 MeV
D
17.28 MeV
Open in App
Solution
Verified by Toppr
Since, p+3Li7→22He4
Thus, proton energy = BE of He atom − BE of Li of atom.
or Ep=(2×4×7.06)−(7×5.60)=17.28MeV where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.
Was this answer helpful?
17
Similar Questions
Q1
If the binding energy per nucleon in 73Li and He nuclei are 5.60MeV and 7.06MeV respectively, then in the reaction p+37Li→24 He, binding energy is
View Solution
Q2
If the binding energy per nucleon in 73Li and 42He nuclei are 5.60MeV and 7.06MeV respectively, then in the reaction : p+73Li→242He energy of proton must be
View Solution
Q3
The binding energy per nucleon of 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 73Li+11H→42He+42He+Q, the value of energy Q released is
View Solution
Q4
If the binding energy per nucleon in 73Li and 42He nuclei are 5.60MeV and 7.06MeV respectively, then in the reaction p+73Li→242He Energy of proton must be -
View Solution
Q5
The Binding energy per nucleon of 73Li and 42He nucleon are 5.60MeV and 7.06MeV, respectively. In the nuclear reaction 73Li+11H→42He+42He+Q, the value of energy Q released is