A hydrogen atom in the ground state absorbs an ultraviolet radiation photon of wavelength 50 nm . Assuming the entire photon energy is taken up by the electron, determine the kinetic energy of the ejected electron.

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Correct answer: A. 11.24 eV

The energy of the incident photon is given by

E = \dfrac{hc}{\lambda}

Using

hc \approx 1242 \text{ eV} \cdot \text{nm}

and

\lambda = 50 \text{ nm}

:

E = \dfrac{1242}{50} = 24.84 \text{ eV}

The binding energy of an electron in the ground state of a hydrogen atom is

13.6 \text{ eV}

.

Using the photoelectric equation, the kinetic energy of the ejected electron is:

K = E - W

K = 24.84 \text{ eV} - 13.6 \text{ eV} = 11.24 \text{ eV}

A hydrogen atom in the ground state absorbs an ultraviolet...

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