Chemistry · Dual nature of matter, de Broglie's relationship, Heisenberg uncertainty principle
An electron in -atom in ground state absorbs 1.5 times as much energy as the min
An electron in \( H \) -atom in ground state absorbs 1.5 times as much energy as the minimum required for its escape (i.e., \( 13.6 e V) \) from the atom. Calculate the wavelength of emitted electron.
- A. \( \lambda=4.70 \times 10^{-10} \mathrm{m} \)
- B. \( \lambda=61 \times 10^{-9} m \)
- C. \( \lambda=4.70 \times 10^{-14} \mathrm{m} \)
- D. \( \lambda=2.7 \times 10^{-10} \mathrm{m} \)
Step-by-step solution
The absorbed energy is 1.5 × 13.6 eV = 20.4 eV. The ionization energy is 13.6 eV, so the kinetic energy of the emitted electron is 20.4 - 13.6 = 6.8 eV. Using de Broglie wavelength λ = h / √(2mK), we compute λ ≈ 4.70 × 10^{-10} m.
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