Question

I bombard a hydrogen atom with electrons and it gets excited tot he n=6, l=1, m_1=-1, m_s=+1/2 energy level. In our usual spectroscopic notation, this is 6p orbital.

A. What is the longest wavelength photon that the atom could emit?

B. What is the shortest wavelength photon that the atom could emit?

Answer 1

Given

   hydrogen atom excited to

n=6, l=1, m_1=-1, m_s=+1/2 energy level.

we know that thechange in energy of the wave when the electron excited from one energy state to the other (ni,nf) is

   delta E = 13.6 eV(1/ni^2 - 1/nf^2)

let the ni = 6 , nf = 5

   E1 = 13.6(1/36-1/25) = -0.16622 eV

ni = 6 , nf = 4

   E2 = 13.6(1/36-1/16) = -0.4722 eV

ni = 6 , nf = 3

   E3 = 13.6(1/36-1/9) = -1.13333 eV

and

ni = 6 , nf = 1

   E4 = 13.6(1/36-1/1) = -13.22 eV   

we know that E = h*nue = hc/Lambda ==> lambda = hc/E

from the above data the transition from ni = 6 to nf = 1, the shortest wavelength photon , will be emitted by the atom

               Lambda = (6.625*10^-34*3*10^8)/(-13.2222*(-1.6*10^-19))= 9.3947111675818*10^-8 m

and for the transitionn from ni = 6 ,nf = 5 , the longest wavelength of the photon .

                       Lambda = (6.625*10^-34*3*10^8)/(-0.16622*(-1.6*10^-19)) = 7.4731530501745*10^-6 m