The wavelength is 91.5 pm ( 91.5 Pico meter).
Explanation:
The formula can be expressed below for electron’s energy,
[tex]\text {Energy of electron}=\frac{p^{2}}{2 m}[/tex]
Where,
p = momentum
m= mass of electron
We know, mass of electron = [tex]9.1 \times 10^{-31} \mathrm{kg}[/tex]
Energy of electron, [tex]1 e V=1.6 \times 10^{-19} \mathrm{J}[/tex]
Therefore, [tex]\text { energy of electron, 180 eV }=180 \times 1.6 \times 10^{-19} J[/tex]
By substituting the known values in the equation, we get,
[tex]180 \times 1.6 \times 10^{-19}=\frac{p^{2}}{2 \times 9.1 \times 10^{-31}}[/tex]
[tex]p^{2}=180 \times 1.6 \times 10^{-19} \times 2 \times 9.1 \times 10^{-31}[/tex]
[tex]p^{2}=5241.6 \times 10^{-50}[/tex]
Taking square root, we get
[tex]\text {Momentum, } p=72.399 \times 10^{-25} \mathrm{kg} . \mathrm{m} / \mathrm{s}[/tex]
We know,
[tex]\lambda=\frac{h}{p}[/tex]
Here, h – Planck constant = [tex]6.626 \times 10^{-34} \mathrm{J.s}[/tex]
So, the wavelength would be,
[tex]\lambda=\frac{6.626 \times 10^{-34}}{72.399 \times 10^{-25}}=0.0915 \times 10^{-34+25}=0.0915 \times 10^{-9} \mathrm{m}[/tex]
Adding [tex]10^{-3}[/tex] in both numerator and denominator we get the value as
[tex]\lambda=0.0915 \times 10^{-9} \times \frac{10^{-3}}{10^{-3}}=0.0915 \times 10^{3} \times 10^{-12}=91.5 \mathrm{pm}[/tex]
Where, pm – Pico meter - [tex]10^{-12}[/tex]
