Atoms - Result Question 14
15. Two particles of masses $m_1, m_2$ move with initial velocities $u_1$ and $u_2$. On collision, one of the particles get excited to higher level, after absorbing energy $\varepsilon$. If final velocities of particles be $v_1$ and $v_2$ then we must have
[2015]
(a) $\frac{1}{2} m_1 u_1^{2}+\frac{1}{2} m_2 u_2^{2}=\frac{1}{2} m_1 v_1^{2}+\frac{1}{2} m_2 v_2^{2}-\varepsilon$
(b) $\frac{1}{2} m_1 u_1^{2}+\frac{1}{2} m_2 u_2^{2}-\varepsilon=\frac{1}{2} m_1 v_1^{2}+\frac{1}{2} m_2 v_2^{2}$
(c) $\frac{1}{2} m_1^{2} u_1^{2}+\frac{1}{2} m_2^{2} u_2^{2}+\varepsilon=\frac{1}{2} m_1^{2} v_1^{2}+\frac{1}{2} m_2^{2} v_2^{2}$
(d) $m_1^{2} u_1+m_2^{2} u_2-\varepsilon=m_1^{2} v_1+m_2^{2} v_2$
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Answer:
Correct Answer: 15. (b)
Solution:
- (b) By law of conservation of energy,
$K . E_f=K . E_i-$ excitation energy $(\varepsilon)$
or $\frac{1}{2} m v_1^{2}+\frac{1}{2} m v_2^{2}=\frac{1}{2} m_1 u_1^{2}+\frac{1}{2} m_2 u_2^{2}-\varepsilon$
