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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2020
PhysicsEasysingle choice

A radioactive material decays by simultaneous emission of two particles with half-lives1620 yrand810 yrrespectively. The time (inyr) after which one-fourth of the material remains, is

Options:

Answer:
D
Solution:

From Rutherford-Soddy law, the number of atoms left afternhalf-lives is given by

         N=N012n

Where,N0is the original number of atoms.

The number of half-life

          n=time of decayeffective half-life

Relation between effective disintegration constant (λ) and half-life (T)is
               λ=ln⁡2T

∴  λ1+λ2=ln⁡2T1+ln⁡2T2

Effective half -life

                 1T=1T1+1T2=11620+1810

                1T=1+21620⇒T⇒540 yr

∴           n= t540

⇒t540=2

⇒      t= 2 × 540 = 1080 yr

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity
⏱ 2mℹ️ Source: PYQ_2020

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