IIT JEE , AIEEE Forum - Physics , Modern Physics

rakesh61

A nucleus of mass number 220, initially at rest, emits an

-particle . If the Q value of the reaction is 5.5 MeV, the energy of the emitted

-particle will be
(a)4.8 MeV
(b)5.4 MeV
(c)6.0 MeV
(d)6.8 MeV

Ans b

okease tell me how to solve this what is Q value in this

rates assured

ramyani

X (220) -> Y ( 216 ) + alpha ( 4 )

nucleus of mass number 220, initially at rest. So conservation of momentum gives

0 = 216 v₁ + 4 v₂

v₁ = - 4v₂ / 216 = -v₂ / 54

now conservation of energy gives

( 1/2) m₁ v_{1² +}( 1/2) m₂ v_{2² = Q

108}v_{1² + 2}v₂ ² = 5.5 mev

108 ( -v₂ / 54 )² + 2 v₂ ² = 5.5 mev

2 v₂ ² = 5.4 Mev

** edited.

ramyani

to solve this Q , 1st what is Q value ?

Allow me to qoute goiit study material :

Nuclear fission:-
(1) The phenomenon of breaking a heavy nucleus in to two light nuclei of almost equal masses along with the release of huge amount of energy is called nuclear fission.
(2) The heavy atoms like uranium (which low binding energy per nucleon) undergo fission reaction to produce two stable nucleii and generate mass defect which convers in to energy
e.g

Dumb Question:
in which form is the energy released.
ans: Energy is released in the form of kinetic energy of fission fragments

Now if that be the case,

In the reaction

X (220) -----> Y ( 216 ) + alpha ( 4 )

The energy is released in the form of KE of fission fragments.

( Plz note that I m only using the concept in case of radioactivity)

nucleus of mass number 220, initially at rest. So conservation of momentum gives

0 = 216 v₁ + 4 v₂

v₁ = - 4v₂ / 216 = -v₂ / 54

now conservation of energy gives

( 1/2) m₁ v_{1² +}( 1/2) m₂ v_{2² =}Q
216 v_{1² +}4 v₂ ² = 5.5 mev

216( -v₂ / 54 )² + 4 v₂ ² = 5.5 mev

4 v₂ ² = 5.4 Mev

The mass no of alpha is 4. so we keep it as it is. Or we can keep m₂ for it which will give same result.

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