SATHEE: Atoms - Result Question 14

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 $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ then we must have

[2015]

(a) $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - ε$

(b) $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

(c) $\frac{1}{2} m_{1}^{2} u_{1}^{2} + \frac{1}{2} m_{2}^{2} u_{2}^{2} + ε = \frac{1}{2} m_{1}^{2} v_{1}^{2} + \frac{1}{2} m_{2}^{2} v_{2}^{2}$

(d) $m_{1}^{2} u_{1} + m_{2}^{2} u_{2} - ε = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}$

Show Answer

Answer:

Correct Answer: 15. (b)

Solution:

(b) By law of conservation of energy,

$K . E_{f} = K . E_{i} -$ excitation energy $(ε)$

or $\frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε$

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