Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, [tex]4.56\times 10^{-19}J[/tex]
Explanation : Given,
Wavelength = [tex]435.8nm=435.8\times 10^{-9}m[/tex]
conversion used : [tex]1nm=10^{-9}m[/tex]
Formula used :
[tex]E=h\times \nu[/tex]
As, [tex]\nu=\frac{c}{\lambda}[/tex]
So, [tex]E=h\times \frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\lambda[/tex] = wavelength = [tex]435.8\times 10^{-9}m[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get:
[tex]E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}[/tex]
[tex]E=4.56\times 10^{-19}J[/tex]
Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, [tex]4.56\times 10^{-19}J[/tex]
