A photon strikes an isolated hydrogen atom in its ground state, ejecting an electron. The de Broglie wavelength of this ejected electron is found to be exactly equal to the de Broglie wavelength of an electron revolving in the first orbit ( n=1 ) of a He ⁺ ion. The energy of the incident photon is ______ eV.

Step-by-step solution

The de Broglie wavelength of an electron is related to its momentum by = h p , and to its kinetic energy by = h 2mK . Since the ejected electron has the same de Broglie wavelength as an electron in the first orbit of a He ⁺ ion, their kinetic energies must be equal. The kinetic energy of an electron in the n^ th orbit of a hydrogen-like species is given by K = 13.6 Z^2 n^2 eV. For the first orbit ( n=1 ) of He ⁺ ( Z=2 ): K = 13.6 2^2 1^2 = 54.4 eV. Thus, the kinetic energy of the ejected electron is 54.4 eV. The en