Respuesta :
Answer : The energy change occurring in the atom due to this emission is, [tex]18.38\times 10^{-26}J[/tex]
Solution :
Formula used :
[tex]\Delta E=\frac{h\times c}{\lambda}[/tex]
where,
[tex]\Delta E[/tex] = change in energy of photon = ?
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]2.998\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = 1.08 m
Now put all the given values in the above formula, we get the energy of the photons.
[tex]\Delta E=\frac{(6.626\times 10^{-34}Js)\times (2.998\times 10^8m/s)}{1.08m}[/tex]
[tex]\Delta E=18.39\times 10^{-26}J[/tex]
Therefore, the energy change occurring in the atom due to this emission is, [tex]18.38\times 10^{-26}J[/tex]
