Answer:
The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹ and λ = 88.31 nm.
Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.
E = hc/λ (1)
Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.
But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.
So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.
Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:
λ = hc/E (2)
Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s
But the Planck constant can be expressed in electron volts:
1 eV = 1.602 x 10⁻¹⁹ J
h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s
h= 4.136x10⁻¹⁵ eV.s
Now, it is convenient to express the speed of light in nanometers:
1nm = 1x10⁻⁹ m
c = 3.00x10⁸ m/ 1x10⁻⁹ m
c = 3x10¹⁷ nm/s
Substituting in equation 2:
λ = (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV
λ = 1240 eV. nm/ 14.04 eV
λ = 88.31 nm
The frenquency is calculated using the equation 2 in the following way:
E = hν (3)
Where ν is the frecuency
ν = E/h
ν = 14.04 eV/4.136×10⁻¹⁵ eV.s
ν = 3.394×10¹⁵ s-1
So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.
