Answer to Question #138600 in Electric Circuits for mungaster

Question #138600
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 nn in the stationary state of hydrogen atom is defined as En=m×e432×π2×ε02×2×1n2E_n=\frac{m\times e^4}{32\times \pi^2\times\varepsilon_0^2\times \hbar^2 } \times \frac {1}{n^2} , where mm is the electron mass, ee the electron charge, ε0\varepsilon_0 – electric constant, \hbar Planck's constant Dirac.

Express nn and you get n=m×e432×π2×ε02×h2×1En=n=\sqrt{\frac{m\times e^4}{32\times \pi^2\times\varepsilon_0^2\times h^2 } \times \frac {1}{ E_n }}= 9.1×1031×(1.6×1019)432×π2×(8.85×1012)2×(1.05×1034)2×12.28×1.6×10192\sqrt{\frac{9.1\times 10^{-31}\times (1.6\times 10^{-19})^4}{32\times \pi^2\times(8.85\times10^{-12})^2\times (1.05\times 10{-34})^2 } \times \frac {1}{ 2.28\times 1.6\times 10^{-19}}}\approx2 .

B) Since the kinetic energy is E=m×v22E=\frac{m\times v^2}{2} ,hence v=2×2.28×1.6×10199.1×1031=8.95×105v=\sqrt{\frac{2\times 2.28\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}=8.95\times 10^5 m/s.


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