Respuesta :
Explanation:
The given data is as follows.
Energy of radiation absorbed by the electron in hydrogen atom = [tex]1.08 \times 10^{-17} J[/tex]
As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.
E = [tex]h \nu[/tex]
[tex]1.08 \times 10^{-17} J[/tex] = [tex]6.626 \times 10^{-34} Js \times \nu[/tex]
[tex]\nu[/tex] = [tex]0.163 \times 10^{17} s^{-1}[/tex]
or, [tex]\nu[/tex] = [tex]1.63 \times 10^{16} s^{-1}[/tex]
It is known that, [tex]\nu = \frac{c}{\lambda}[/tex]
[tex]1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}[/tex]
[tex]\lambda[/tex] = [tex]1.84 \times 10^{-8} m[/tex]
And, according to De-Broglie equation [tex]\lambda = \frac{h}{p}[/tex]
as, p = [tex]m \times \nu[/tex]
So, [tex]\lambda = \frac{h}{m \times \nu}[/tex]
[tex]m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}[/tex]
= [tex]3.6 \times 10^{-26} J/m[/tex]
Now, on squaring both the sides we get the following.
[tex](m \times \nu)^{2}[/tex] = [tex](3.6 \times 10^{-26} J/m)^{2}[/tex]
= [tex]12.96 \times 10^{-52}[/tex]
[tex]m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}[/tex]
where, m = mass of electron
So, [tex]m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}[/tex]
= [tex]\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}[/tex]
= [tex]1.42 \times 10^{-21}[/tex] J
Since, K.E = [tex]\frac{1}{2}m \nu^{2}[/tex]
= [tex]\frac{1.42 \times 10^{-21} J}{2}[/tex]
= [tex]0.71 \times 10^{-21} J[/tex]
Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is [tex]7.1 \times 10^{-22} J[/tex].
