Question

A stationary hydrogen atom deexcites from first excited state to ground state. Find recoil speed of hydrogen atom up to nearest integer value. 
(mass of hydrogen atom = \(1.8 \times 10^{-27}\) \(kg\))

Answer 1

The absolute change in ground state energy is \(|ΔE_0| = (-13.6 \left\{1-\frac{1}{4}\right\} )ev\). The absolute change in energy is \(|ΔE| = 10.2\ ev\). The wavelength is calculated as \(λ=\frac{12400}{10.2} × 10^{-10}\ m\). Momentum is given by \(ρ=\frac{h}{λ} \), which equals \(\frac{6.63 × 10^{-34} × 10.2}{12400 × 10^{-10}}\). Since \(mv=\frac{h}{λ}\), the value is \(1.8 × 10^{-27}\). The velocity is \(v = \frac{6.63×10.2×10^{-34}}{12400 × 10^{-10}}\), which simplifies to \(v =\frac{6.63 × 10.2}{12400 × 1.8} × 10^3\), further reducing to \(\frac{6.63 × 102}{124 × 1.8}\), and finally \(3.02 ≈ 3\ m/s\). Therefore, the recoil speed of the hydrogen atom is approximately 3 m/s.