What is the change in energy if the electron from part a now drops to the ground state?what is the wavelength λ of the photon that has been released in part b?

Respuesta :

Change in energy: ΔE = E1 - E4 


E1 = -13.6 eV(1.602 x 10^-19 J/eV)/1^2 = -2.18 x 10^-18 J 


ΔE = -2.18 x 10^-18 J - (-1.36 x 10^-19 J) = -2.04 x 10^-18 J 



Wavelength of the photon: E = hν = hc/λ, where h = Planck's constant and v = frequency in s^-1; λ = wavelength in m, and c = 3.00 x 10^8 m/s 


2.04 x 10^-18 J = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s)/λ 


λ = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s)/(2.04 x 10^-18 J) = 9.74 x 10^-8 m