Question

The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of photons capable of just ionizing hydrogen atoms? values for constants can be found here.

Answer 1

1) Frequency: $3.29\cdot 10^{15}Hz$

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

$E=2.18 aJ=2.18\cdot 10^{-18} J$

The energy of a photon is given by

$E=hf$

where $h=6.63\cdot 10^{-34}Js$ is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

$f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz$

2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

$\lambda=\frac{c}{f}$

where $c=3\cdot 10^8 m/s$ is the speed of light and f is the frequency. Substituting the frequency, we find

$\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm$