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A proton is bombarded on a stationary li...

A proton is bombarded on a stationary lithium nucleus. As a result of the collision, two `alpha`-particles are produced. If the direction of motion of the `alpha`-particles with the initial direction of motion makes an angles `cos^-1(1//4)`, find the kinetic energy of the striking proton. Given, binding energies per nucleon of `Li^7` and `He^4` are 5.60 and 7.06 MeV, respectively.
(Assume mass of proton `~~` mass of neutron).

Text Solution

Verified by Experts

The correct Answer is:
A, B
Q-value of the reaction is
`Q=(2xx4xx7.06-7xx5.6)MeV=17.28MeV`
Applying conservation of energy for collision.
`K_p+Q=2Ka` …(i)
(Here, `K_p ` and `K_alpha` are the kinetic energies of proton and `alpha`-particle respectively)

From conservation of linear momentum,
`sqrt(2m_pK_p)=2sqrt(2m_(alpha)K_(alpha))costheta` ...(ii)
:. `K_p=16K_alphacos^2theta=(16K_alpha)(1/4)^2` (as `m_(alpha)=4m_p`)
:. `K_(alpha)=K_p` ...(iii)
Solving Eqs. (i) and (iii) with `Q=17.28MeV`
We get `K_p=17.28MeV`
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