Saved Bookmarks
| 1. |
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength andfrequency of photon. |
|
Answer» Solution :Energy of ELECTRON in `n^(th)` orbit of atom `E_(n) = - (13.6)/(n^(2))eV` `rArr`The energy of electron in ground state `E_(1)=-(13.6)/((1)^(2))eV [ :.` For ground state n=1] `:.E_(1)=-13.6eV` `rArr` Energy of electron in n=4 orbit, `E_(4)=-(13.6)/((4)^(2))eV` `=-(13.6)/(16)eV` `:.E_(4)=-0.85ev` `rArr` DIFFERENT in enegy of electron, `DeltaE=E_(4)-E_(1)` `=-0.85eV-(13.eV)` `=12.75 eV` br> `=12.75xx1.6xx10^(-19)J` `=20.4xx10^(-19)J` `rArr` Now, `DeltaE=hv` `:.v=(DeltaE)/(h)=(20.4xx10^(-19))/(6.625xx10^(-34))` `:.v=3.073xx10^(15)Hz` `rArr` , now `C= lambda v` `:. lambda (c)/(v)=(3xx10^(8))/(30.79xx10^(15))` `:. lambda=97.43xx10^(-9)=97.43 nm` |
|