Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Under certain circumstances, a nucleus can decay by emitting a particle more massive than an alpha- particle. Consider the following decay processes. Calculate the Q -values for these decays and determine that both are energetically allowed.

Q.13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an \alpha-particle. Consider the following decay processes:

                  _{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}

                  _{88}^{223}\textrm{Ra}\rightarrow _{86}^{219}\textrm{Rn}+_{2}^{4}\textrm{He}

Calculate the Q-values for these decays and determine that both are energetically allowed.

Answers (1)

best_answer

_{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}

\\\Delta m=m(_{88}^{223}\textrm{Ra})-m(_{82}^{209}\textrm{Pb})-m(_{6}^{14}\textrm{C})\\ =223.01850-208.98107-14.00324 \\=0.03419u

1 u = 931.5 MeV/c2

Q=0.03419\times931.5

=31.848 MeV

As the Q value is positive the reaction is energetically allowed

_{88}^{223}\textrm{Ra}\rightarrow _{86}^{219}\textrm{Rn}+_{2}^{4}\textrm{He}

\\\Delta m=m(_{88}^{223}\textrm{Ra})-m(_{86}^{219}\textrm{Rn})-m(_{2}^{4}\textrm{He})\\ =223.01850-219.00948-4.00260 \\=0.00642u

1 u = 931.5 MeV/c2

Q=0.00642\times931.5

=5.98 MeV

As the Q value is positive the reaction is energetically allowed

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question