Question

If the binding energy per nucleon in Li and He nuclei are 5.60 MeV and 7.06 MeV respectively then in the reaction:
p+ Li--------->2 He
Energy of proton must be

Answer 1

@michele_laino sir plss..help

Answer 2

do we know the energy or the so called \(Q\) of such reaction?

Answer 3

Nope it is not mentioned....

Answer 4

please wait I make some computations...

Answer 5

Okk...i will...

Answer 6

if I use the connservation of the four-momentum, I get this:
\[\Large {E_p} = \frac{{4{{\left( {{M_{He}}{c^2}} \right)}^2} - {{\left( {{M_p}{c^2}} \right)}^2} - {{\left( {{M_{Li}}{c^2}} \right)}^2}}}{{2{M_{Li}}{c^2}}}\]
where \(E_p\) is the total energy of the proton

Answer 7

Sorry sir...bt i didn't get u...this formula
From where it comes..??

Answer 8

it comes from the conservation of the four-momentum, namely we have to consider the vectors like objects of the Minkowski space

Answer 9

Four-momentum???

Answer 10

please what are your options?

Answer 11

28.24
17.28
1.46
39.2
all r in MeV

Answer 12

I'm thinking...

Answer 13

Okk...

Answer 14

my formula above is wrong

Answer 15

So what now ?

Answer 16

please wait...

Answer 17

So sorry....sir.

Answer 18

no problem :)
the product of such reaction are two atoms of Helium?

Answer 19

Yep...

Answer 20

the number of mass is not conserved:
\[\Large _1^1{\text{p}} + _3^6{\text{Li}} \to _2^4{\text{He}} + _2^4{\text{He}}\]

Answer 21

Lithium's isotope of mass number 7

Answer 22

as we can see at the left we have \(1+6=7\), whereas at the right we hae \(4+4=8\)
ok!

Answer 23

Sorry didn't mention in the ques itself. ...

Answer 24

then we have a reaction like this:
\[\Large _1^1{\text{p}} + _3^7{\text{Li}} \to _2^4{\text{He}} + _2^4{\text{He}}\]

Answer 25

Yo exactly

Answer 26

I think that we can apply the conservation of energy like in the photoelectric effect, so we can write:
\[\huge E + 5.60 = 7.06\]

Answer 27

Bt there is some release in energy whenever a nuxlear reaction takes place

Answer 28

How can we apply energy conservation here?

Answer 29

yes! sure there is the so called \(Q\) of the reaction

Answer 30

I think that \(E\) stands for kinetic energy only of the proton

Answer 31

Then here no Q is mentioned then how can v proceed to the problem

Answer 32

Sorry sir it's 2:17 am here.....i need to go...
Feeling very sleepy...plss..wud u help me vid dis problem l8r....

Answer 33

the only idea that I have in my mind is the equation above. Otherwise we have to consider the conservation of four-momentum, nevertheless, as I wrote before such conservation gives a wrong equation, since don't know the \(Q\) of the reaction

Answer 34

ok! I will help you!

Answer 35

given the reaction above, we can write this energy balance:
\[\Large {M_p}{c^2} + {M_{Li}}{c^2} = 2{M_{He}}{c^2} + Q\]
where \(Q\) is the energy freed by such reaction like gamma rays, otherwise such reaction is not possible

Answer 36

Yess correct we can write that since Q value is equal to the
(mass of reactants-mass of product)*C^2

Answer 37

Bt sir how can we assume Q value to be 0 unless mentioned....?

Answer 38

after a substitution, we get:
\[\Large 940 + 5.6 = \left( {2 \times 7.06} \right) + Q\]
and we are able to compute the value of \(Q\)

Answer 39

What is 940 ??

Answer 40

the mass of proton, is:
\[\Large {M_p}{c^2} = 940\;{\text{MeV}}\]

Answer 41

Okk 931 MeV r u saying that???

Answer 42

1 amu=931 MeV

Answer 43

yes! I have rounded the value

Answer 44

Ohh well...it's alryt...then...so do u want to say that energy of proton is 931 MeV in this question?

Answer 45

yes!

Answer 46

I mean is the ans according to u 940 MeV ?

Answer 47

Sorry bt that is not in the options ....

Answer 48

no it is not, since it is the energy balance in order to make such reaction

Answer 49

So after finding out the Q value of the reaction what shall we do ....n even v have calculated the q value by supposing the proton enrgy to be 940 MeV now how can we change it jst for the sake of ans...

Answer 50

please wait, I try to make some computations...

Answer 51

Okay sir!!!

Answer 52

I'm sorry I got a wrong result.

Answer 53

I go offline to review my notes on nuclear physics, and I try to solve such problem

Answer 54

It"zz okkay with me sir....

Answer 55

1H + 7Li ® 4He + 4He

B.E.(1)= 0.0 ± 0.0 keV
B.E.(2)= 39244.526 ± 0.473 keV
B.E.(3)= 28295.673 ± 0.005 keV
B.E.(4)= 28295.673 ± 0.005 keV
Q-value: 17346.82 keV
Uncertainty: 0.473 keV (ignoring correlations)

Threshold: 0 keV

in order to make the reaction happen, we need energy of Q-value. I do not know whether 17.34 MeV correct. What is the answer?

Answer 56

here are my computations:
the masses of nuclide are:
\[\begin{gathered}
{M_{Li}} = \left( {3 \times 938 + 4 \times 939} \right) - \left( {7 \times 5.6} \right) = 6530.8\;{\text{MeV}} \hfill \\
\hfill \\
{M_{He}} = \left( {2 \times 938 + 2 \times 939} \right) - \left( {4 \times 7.06} \right) = 3725.76\;{\text{MeV}} \hfill \\
\hfill \\
{M_p} = 938\;{\text{MeV}} \hfill \\
\hfill \\
{M_n} = 939\;{\text{MeV}} \hfill \\
\end{gathered} \]
from the conservation of the four-momentum in the center of mass of products, the minimum kinetic energy of the proton, is:
\[{E_p} = \frac{{4{{\left( {{M_{He}}} \right)}^2} - {{\left( {{M_p}} \right)}^2} - {{\left( {{M_{Li}}} \right)}^2}}}{{2{M_{Li}}}} \cong 917\;{\text{MeV}}\]
and the \(Q\) of such reaction, is:
\[938 + 6530.8 = 2 \times 3725.76 + Q \Rightarrow Q = 17.28\;{\text{MeV}}\]

Answer 57

@Michele_Laino sir the ans u r cuming up with as Q value here is the ans for the energy of proton....in the ans key...

Answer 58

Ans is 17.28 MeV

Answer 59

yes! I think so!

Answer 60

Okk...sir bt we can use ur previous method also...where we can use direct formula for Q value as
Binding energy of helium- binding energy of lithium ....

Answer 61

I think that both formulas are equivalent

Answer 62

From this we will get the Q value of the reaction n as u proposed to me that this Q value is what will be the kinetic energy of proton...so final ans is this only...

Answer 63

yes! Since I think that such \(Q\) value can be attributable like kinetic energy of the proton

Answer 64

Sir is it always the case ...like whatsoever is the enrgy released in a reaction would be attributed to some particles...

Answer 65

for example, let's suppose to consider this reaction:
\[\Large \gamma \to {e^ + } + {e^ - }\]
namely the pair creation

Answer 66

the minimum energy of the products is \(0.51 \times 2=1.02\) MeV, when electron and positron are at the rest

Answer 67

in order to make such reaction, the minimum energy of gamma ray, has to be just 1.02 MeV, otherwise such reaction doesn't happen

Answer 68

Okk so gamma particle will have to move with the energy mentioned in order to let go the reaction to its good pace....

Answer 69

yes!

Answer 70

Similarly in this case for reaction to occure we will ensure that the Q value goes to the enrgy of proton in order for completion of reaction....

Answer 71

here the rest mass of the electron or the rest mass of positron are 0.51 MeV

Answer 72

yes! that's right!

Answer 73

Sir the Q value here will be negative in case of tge example u took....

Answer 74

The*

Answer 75

subsequently part of the total amount of energy goes for creation of the products namely the two atoms of helium, and the remaining part goes for gamma rays and recoil energy of the two atoms of helium

Answer 76

Bcoz the Q value =
(mass of reactants- mass of products)*c^2 so will give us a negative value ..
Isn't it...?

Answer 77

yes! this is not a contradiction. It means that part of the energy of the proton will be used in order to create the products

Answer 78

Okk so u mean that Q value will be 0 if we just take the energy of proton to be 17.28 MeV n that's how it has to be done...

Answer 79

Or better if we consider ur example only then energy of gamma particle should be such that Q value turns out to be just 0
Bcoz Q value -ve implies reaction will not occur

Answer 80

no, please I say that the \(Q\) value, is the minimum energy, kinetic energy since we have a free proton, which such proton have to have in order to make the reaction

Answer 81

I say free proton, since that proton is moving inside a space without any field of force

Answer 82

Sir was my statement wrong???
Why can't it be in that manner since we know that Q value has to be a positive quantity always for reaction to b feasible then we may consider the extreme case if it...n then solve the problem

Answer 83

total energy \(E\) of the proton, measured with respect to the laboratory system, is:
\[\Large E = \left( {T + m} \right){c^2} = 17.28 + 938 = 955.28\;{\text{MeV}}\]

Answer 84

the \(Q\) of a reaction can be a negative number

Answer 85

BT then the reaction won't proceed

Answer 86

no, no, if we have a reaction with negative \Q\), it means that we have to add energy to the system

Answer 87

oops.. negative \(Q\)*...

Answer 88

Sorry sir ....It was given in one of the books I read....maybe it says that the reaction wouldn't be feasible thereby

Answer 89

I think that it is not spontaneous

Answer 90

Yupp.....that's what I meant

Answer 91

So u were saying something about conservation

Answer 92

yes!
nevertheless if we add energy to the system, such reaction can be feasible

Answer 93

Feasible was the blunder .......lol

Answer 94

ok! :)

Answer 95

here is another example from my nuclear physics textbook

Answer 96

Yes ...I m here

Answer 97

U can carry on..

Answer 98

\[\huge n + p = d + 2.225\]
neutron and proton are at the rest, and they combine in order to make a deuteron, with \(Q=2.225\) MeV