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=(E)/(h)=(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=(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=(3\cdot 10^8 m/s)/(3.29\cdot 10^(15)Hz)=9.12\cdot 10^(-8) m=91.2 nm`