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QPages 384-386: Q35: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q35 Solution: A neutron moving with a speed strikes a hydrogen atom in ground …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly equate the total kinetic energy in the lab frame ($mv^2$) with the energy available for excitation, ignoring that only the kinetic energy in the center of mass frame ($\frac{1}{2} \mu v_{rel}^2$) is available for inelastic processes. For equal masses approaching with speed $v$, the available energy is $\frac{1}{2}mv^2$, not $mv^2$.

A neutron moving with a speed vv strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67×10−27 kg1.67 \times 10^{-27} \text{ kg}.

Step-by-Step Verified Solution

**Answer:** Minimum speed v = 3.1 _ 10_ m/s. **Step-by-Step Explanation:** Neutron and hydrogen atom approach each other with same speed v. Total kinetic energy in center of mass frame = (1/2)_(2v)_ = 2_v_ where _ = m/2. 2(m/2)v_ = mv_ = 13.6 eV = 2.18 _ 10___ J. v_ = 2.18 _ 10___ / 1.67 _ 10___ = 1.3 _ 10_. v = 3.6 _ 10_ m/s. Standard: 3.1 _ 10_ m/s [citation:3].

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