A photon with energy 2.28 eV is absorbed by a hydrogen atom. Find (a) the minimum n for a hydrogen atom that can be ionized by such a photon and (b) the speed of the electron released from the state in part.
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Expert's answer
2020-10-19T13:20:40-0400
A) since the energy of level n in the stationary state of hydrogen atom is defined as En=32×π2×ε02×ℏ2m×e4×n21 , where m is the electron mass, e the electron charge, ε0 – electric constant, ℏ Planck's constant Dirac.
Express n and you get n=32×π2×ε02×h2m×e4×En1=32×π2×(8.85×10−12)2×(1.05×10−34)29.1×10−31×(1.6×10−19)4×2.28×1.6×10−191≈2 .
B) Since the kinetic energy is E=2m×v2 ,hence v=9.1×10−312×2.28×1.6×10−19=8.95×105 m/s.
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