A radioactive element decays by $β$ -emission. If the mass of parent and daughter nuclide are $m_{1}$ and $m_{2}$ respectively, the energy liberated during the emission is :

[m_{1} - m_{2} - 2 m_{e}] c^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[m_{1} - m_{4} - 2 m_{e}] c^{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[m_{1} - m_{1} - 3 m_{e}] c^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $[m_{1} - m_{2} - 2 m_{e}] c^{2}$
The masses of parent and daughter elements nucleus are $m_{1}$ and $m_{2}$ respectively.
Mass of an atom of parent element $= m_{1} + Z m_{e}$
(where Z is atomic. no. of the parent element)
Mass of an atom of daughter element $= m_{2} + (Z + 1) m_{e}$
(As atomic. no. of element increases by one due to $β$ -emission)
Therefore mass change $=$ mass of parent atom $-$ mass of one $β$ particle lost $- [$ Mass of one daughter atom along with one electron $]$
Mass change $= m_{1} + Z m_{e} - m_{e} - [m_{2} + (Z + 1) m_{e}]$
Mass change $= [m_{1} - m_{2} - 2 m_{e}]$
The energy liberated is the product of mass change and the square of the speed of light. It is $[m_{1} - m_{2} - 2 m_{e}] c^{2}$
Here, c is the speed of light.

Suggest Corrections

Similar questions

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is: