Question

The kinetic energy of an alpha particle which flies out of the nucleus of a Ra226 atom in radioactive disintegration is 4.78 MeV. Find the total energy evolved during the escape of the alpha particle.

Answer 1

Answer:

The total energy evolved during the escape of the alpha particle is 4.87 MeV.

Explanation:

Given,

The kinetic energy of an alpha particle which flies out of the nucleus of a Ra226 atom in radioactive disintegration[E(a)] is 4.78 MeV.

To find: the total energy evolved during the escape of the alpha particle.

Solution:

The concept is that the conservation of momentum will be observed as the alpha particle will escape out so there will be recoiling momentum of the daughter nucleus.

So, Momentum of alpha particle  = Momentum of daughter nucleus

We can find out total energy evolved by simply adding the kinetic energy of  alpha particle and Kinetic energy of daughter nucleus.

Realtion between kinetic energy and momentum is:

K.E = p²/ 2m

If m(a) is the mass of alpha particle and m(d) is the mass of daughter nucleus,and, p(a) is the momentum of alpha particle and p(d) is the momentum of daughter nucleus then total energy can be given as:

Total energy = p(a)² / 2m(a) + p(d)² / 2m(d)

Since, momentum will be conserved, p(a) = p(d)

Total energy = p(a)² / 2  * [{1 / m(a) } + {1 / m(d)} ]

or, Total energy = m(a)*E(a)  *  [{1 / m(a) } + {1 / m(d)} ]

or, Total energy = E(a)* [ 1 + m(a) / m(d)]

Substituting the values,

Total energy = 4.78 * [ 1 + 4 /222]

or, Total Energy = 4.87 MeV

Therefore, The total energy evolved during the escape of the alpha particle is 4.87 MeV.

Answer 2

Given

Kinetic energy of alpha particle is Ra 226

Disintegration is 4.78 MeV

To find

The total energy evolved during the escape of the alpha particle

Solution

Mass of alpha particle $m_{a}$ =4m, where m is mass of a proton.

mass of Radium nucleus is $m_{u}$ =226m after emission the residue will have mass $m_{r}$ =226 −4 = 222m

We know KE = momentum^2/2m

so for alpha particle, it should be like this E=pa^2/2ma

​or momentum of alpha particle is  pa = $\sqrt[2]{2maE}$ = $\sqrt[2]{8mE}$

​But during emission the residue nucleus   should get same momentum in opposite direction to make net momentum as zero

before and after emission so it should also be  pr = $\sqrt[2]{8mE}$

If kinetic energy of residue is Er then Er = pr^2/2mr = 2×222m/

8mE = 0.018E

So net energy released in the emission is the sum of above two i.e. E + 0.018E = 1.018E

Hence, E=4.78Mev we get total energy as 4.866Mev