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Question: If the ionization potential for hydrogen atoms is 13.6eV, then the wavelength of light required for ...

If the ionization potential for hydrogen atoms is 13.6eV, then the wavelength of light required for the ionization of Hydrogen atoms would be?

Explanation

Solution

Ionization potential or energy is the minimum energy required to remove the outermost electron or the loosely bound electron from an isolated atom or molecule. It is denoted as IE (Ionization energy).

Complete answer:
We are given the ionization potential for Hydrogen atoms which is equal to 13.6 eV. To convert the term eV into Joules, we will multiply it by 1.6×10−19J1.6 \times {10^{ - 19}}J .
Remember the conversion factor of eV to joules as: 1eV=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J
Therefore 3.6eV=3.6×1.6×10−19J3.6eV = 3.6 \times 1.6 \times {10^{ - 19}}J
Now we know the ionization energy in Joules. To find out the wavelength required we will use the relationship between energy, frequency as wavelength. This is known as the Plank’s equation. It states that the energy of a photon like particle will be equal to the product of its frequency and the Plank’s constant ‘h’. Mathematically it can be given as: E=hνE = h\nu
Where ν\nu is the frequency of the light. We know that frequency and wavelength are inversely proportional. Hence the equation involving wavelength can be given as: E=hν=hcλE = h\nu = \dfrac{{hc}}{\lambda }
Where λ\lambda is the wavelength of light and c is the velocity of light. ‘h’ is the Planck's Constant and has a fixed value of 6.626×10−34Js6.626 \times {10^{ - 34}}Js .
The values known to us are:
E=3.6×1.6×10−19J,h=6.63×10−34Js,c=3×108m/sE = 3.6 \times 1.6 \times {10^{ - 19}}J,h = 6.63 \times {10^{ - 34}}Js,c = 3 \times {10^8}m/s . If the value of wavelength is to be found in cm, then we’ll consider the velocity of light in cm/s which will be 3×1010cm/s3 \times {10^{10}}cm/s .
Substituting the values to find wavelength: λ=hcE\lambda = \dfrac{{hc}}{E}
λ=6.63×10−34×3×1083.6×1.6×10−19=9.12×10−8m\lambda = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.6 \times 1.6 \times {{10}^{ - 19}}}} = 9.12 \times {10^{ - 8}}m
Hence the wavelength of light required for ionization would be 912A∘912\mathop A\limits^ \circ

Note:
Remember that wavelength of light has the SI unit of meters, but commonly represented in Armstrong and nanometres itself. The conversion units are: 1A∘=10−10m1\mathop A\limits^ \circ = {10^{ - 10}}m , 1nm=10−9m1nm = {10^{ - 9}}m
Hence to convert the above meters into Armstrong we get 9.12×10−8m=912×10−10m=912A∘9.12 \times {10^{ - 8}}m = 912 \times {10^{ - 10}}m = 912\mathop A\limits^ \circ .