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JEE Main 202122 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
22 July
Shift
Morning Shift
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2021 · 22 July · Morning Shift · Physics · Nuclear Physics

A nucleus with mass number 184 initially at rest emits an α -...

A nucleus with mass number 184184 initially at rest emits an α−\alpha -particle. If the QQ value of the reaction is 5.5MeV,5 . 5MeV, calculate the kinetic energy of the α−\alpha - particle.A nucleus with mass number 184 initially at rest emits an α - particle. If the Q value of the reaction is 5 . 5MeV,

Options

  1. A5.0MeV5 . 0MeV
  2. B5.5MeV5 . 5MeV
  3. C0.12MeV0 . 12MeV
  4. D5.38MeV5 . 38MeV

Correct answer

D. 5.38MeV5 . 38MeV

Step-by-step solution

A nucleus with mass number 184 initially at rest emits an α - particle. If the Q value of the reaction is 5 . 5MeV,

12(4m)v2+12(180m)(4v180)2=5.5MeV\dfrac{1}{2}\left(4m\right){v}^{2} + \dfrac{1}{2}\left(180m\right){\left(\dfrac{4v}{180}\right)}^{2} = 5 . 5MeV

⇒124mv2[1+45(4180)2]=5.5MeV⇒ \dfrac{1}{2}4{mv}^{2}\left[1 + 45{\left(\dfrac{4}{180}\right)}^{2}\right] = 5 . 5MeV
⇒K.E.α=5.51+45⋅(4180)2MeV=5.38MeV⇒ K . E{.}_{\alpha} = \dfrac{5 . 5}{1 + 45 \cdot {\left(\dfrac{4}{180}\right)}^{2}}MeV = 5 . 38MeV

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