Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
An electron collides with a free molecules initially in its ground state. The collision leaves the molecules in an excited state that is metastable and does not decay to the ground state by radiation. Let $\mathrm{K}$ be the sum of the initial kinetic energies of the electron and the molecule, and $\vec{P}$ the sum of their initial momenta. Let $\mathrm{K}^{\prime}$ and $\vec{P}$ 'represent the same physical quantities after the collision. Then -
PhysicsCenter of Mass Momentum and CollisionKVPYKVPY 2011 (SB/SX)
Options:
  • A $K=K^{\prime}, \vec{P}=\vec{P}^{\prime}$
  • B $K^{\prime} < K, \vec{P}=\vec{P}^{\prime}$
  • C $K=K^{\prime}, \vec{P} \neq \vec{P}^{\prime}$
  • D $K^{\prime} < K, \vec{P} \neq \vec{P}^{\prime}$
Solution:
2884 Upvotes Verified Answer
The correct answer is: $K^{\prime} < K, \vec{P}=\vec{P}^{\prime}$
Collision of e $^{-}$lead to excitation of molecules so Collision is inelastic
$\therefore \mathrm{K}^{\prime} < \mathrm{K}$ and loss of kinetic energy go for excitation of molecules. Momentum remain conserved during collision.
$\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{P}^{\prime}}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.