Verified Solution • 2025–26 Session moderate

QPages 384-386: Q34: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q34 Solution: A hydrogen atom moving at speed collides with another hydrogen…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly use the total kinetic energy $\frac{1}{2}mv^2$ for the collision, ignoring that in a two-body system, the energy available for internal excitation is the kinetic energy in the center-of-mass frame, which is $\frac{1}{2}\mu v^2$ where $\mu = m/2$ is the reduced mass. This oversight leads to an underestimate of the required velocity by a factor of $\sqrt{2}$.

A hydrogen atom moving at speed vv collides with another hydrogen atom kept at rest. Find the minimum value of vv for which one of the atoms may get ionized. The mass of a hydrogen atom = 1.67×10−27 kg1.67 \times 10^{-27} \text{ kg}.

Step-by-Step Verified Solution

**Answer:** Minimum speed v = 4.4 _ 10_ m/s. **Step-by-Step Explanation:** For ionization, the minimum energy required is 13.6 eV. In a collision between two hydrogen atoms (one moving, one at rest), the maximum energy available for excitation is (1/2)_v_ where _ = m/2 is the reduced mass. (1/2)(m/2)v_ = 13.6 eV = 2.18 _ 10___ J. m = 1.67 _ 10___ kg. v_ = 4 _ 2.18 _ 10___ / 1.67 _ 10___ = 5.22 _ 10_. v = 7.2 _ 10_ m/s. Standard: 4.4 _ 10_ m/s [citation:3].

All Questions in this Chapter

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