The quantim mechanical treatment of the hydrogen atom gives the energy vale:E_n=(-13.6)/(n^(2))eV "atom"^(-1)(i) use thios expression to find /_\Ebetween=3 and n=4(ii) Calcuilate the wavelength corresponding to the above transition.

The quantim mechanical treatment of the hydrogen atom gives the energy

1.	The quantim mechanical treatment of the hydrogen atom gives the energy vale:E_n=(-13.6)/(n^(2))eV "atom"^(-1)(i) use thios expression to find /_\Ebetween=3 and n=4(ii) Calcuilate the wavelength corresponding to the above transition.
Answer» Solution :(i) When n=3 `E_3=(-13.6)/(3^(2))=(-13.6)/(9)=-1.511eV "atom"^(-1)` When n=5`E_4=(-13.6)/(4^(2))=(-13.6)/(16)=-0.85 eV "atom"^(-1)` `/_\E=E_4-E_3` `=-0.85-(-1.511)=+0.661eV "atom"^(-1)` `/_\E=E_3-E_4` `=-1.511-(-0.85)` `=-0.661 eV " atom"^(-1)` (ii) WAVE length=`gamma ` `/_\E=(hc)/(gamma)` `therefore gamma=(hc)/(/_\E)` `/_\E=0.661xx1.6xx10^(-19)J` `=1.06xx10^(-19)j` h=planck.s constant=`6.626xx10^(-34)JS^(-1)` `c=3xx10^(8)m/s` `therefore gamma=(6.626xx10^(-34)xx3xx10^(8))/(1.06xx10^(-19))` `=1875xx10^(-6)m.`