Answer to Question #138600 in Electric Circuits for mungaster

Question

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.

Answer 1

A) since the energy of level $n$ in the stationary state of hydrogen atom is defined as $E_n=\frac{m\times e^4}{32\times \pi^2\times\varepsilon_0^2\times \hbar^2 } \times \frac {1}{n^2}$ , where $m$ is the electron mass, $e$ the electron charge, $\varepsilon_0$ – electric constant, $\hbar$ Planck's constant Dirac.

Express $n$ and you get $n=\sqrt{\frac{m\times e^4}{32\times \pi^2\times\varepsilon_0^2\times h^2 } \times \frac {1}{ E_n }}=$ $\sqrt{\frac{9.1\times 10^{-31}\times (1.6\times 10^{-19})^4}{32\times \pi^2\times(8.85\times10^{-12})^2\times (1.05\times 10{-34})^2 } \times \frac {1}{ 2.28\times 1.6\times 10^{-19}}}\approx2$ .

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