Q90 / 90 · KVPY 2011 (SB/SX) · Physics
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 -
- A$K=K^{\prime}, \vec{P}=\vec{P}^{\prime}$
- Correct answerB$K^{\prime} < K, \vec{P}=\vec{P}^{\prime}$
- C$K=K^{\prime}, \vec{P} \neq \vec{P}^{\prime}$
- D$K^{\prime} < K, \vec{P} \neq \vec{P}^{\prime}$
Solution
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.
$\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{P}^{\prime}}$
Hence, the correct answer is option: 'B'.