A free neutron decays to a proton and electron and antineutrino n → ρ + e - + ν ¯ e . Calculate the energy produced in this reaction in MeV . Given m e 0 1 = 9 × 10 - 31 Kg m n 1 0 = 1 . 6747 × 10 - 27 Kg m H 1 1 = 1 . 6725 × 10 - 27 Kg

Mass defect = mass of neutron − (mass of proton + mass of electron) or Δ m = 1 . 6747 × 10 - 27 − ( 1 . 6725 × 10 - 27 + 9 × 10 - 31 ) = 1 . 3 × 10 - 30 kg Energy produced E = Δ m × c 2 = 1 . 3 × 10 - 30 × ( 3 × 10 8 ) 2 ⇒ E = 11 . 7 × 10 - 14 J ⇒ E = 11 . 7 × 10 - 14 1 . 6 × 10 - 19 eV ⇒ E = 7 . 31 × 10 5 eV ⇒ E = 0 . 731 MeV

E M B I B E

Bureau's Higher Secondary Physics Class 12>Chapter 17 - Nuclear Physics>MODEL QUESTIONS

MODEL QUESTIONS

Author:Odisha Board

Odisha Board Physics Solutions for Exercise - MODEL QUESTIONS

Simple step-by-step solutions to MODEL QUESTIONS questions of Nuclear Physics from Bureau's Higher Secondary Physics Class 12. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from MODEL QUESTIONS with Hints & Solutions

EASY

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 17 - Nuclear Physics>MODEL QUESTIONS>Q 19

A free neutron decays to a proton and electron and antineutrino $(n \to ρ + e^{-} + {\bar{ν}}_{e})$ . Calculate the energy produced in this reaction in $MeV$ .

Given

$m ({}_{1}e^{0}) = 9 \times 10^{- 31} Kg m ({}_{0}n^{1}) = 1.6747 \times 10^{- 27} Kg m ({}_{1}H^{1}) = 1.6725 \times 10^{- 27} Kg$