Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# A Free Neutron Beta-decays to a Proton with a Half-life of 14 Minutes. (A) What is the Decay Constant? (B) Find the Energy Liberated in the Process. - Physics

Sum

A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.

(Use Mass of proton mp = 1.007276 u, Mass of ""_1^1"H" atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

#### Solution

Given:
Half-life period of free neutron beta-decays to a proton, T_"1/2" = 14 minutes

Half-life period , T_"1/2" = 0.6931/lambda

Here,  lambda = Decay constant

therefore lambda = 0.693/(14 xx 60)

= 8.25 xx 10^-4  "S"^-1

If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.

therefore "Energy liberated" , E = [m_n - (m_p + m_e)] c^2

= [1.008665  "u" - (1.007276 + 0.0005486) "u"]c^2

= 0.0008404 xx 931  "MeV"

= 782  "keV"

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Q 11 | Page 442