Question

An electron in an hydrogen atom in an second excited state jumps to the first state and ground state of the atom.Find the ratio of wavelength emitted during this process

Answer 1

Acc to me,

I hope You  know the formula

1/W=R{1/(nf)^2-1/(ni)^2}

W=wavelength

nf=final state

ni=intial state

I hope You know for ground state n=1,for 1st excited state n=2,2nd excited state n=3

In question given that electron go from 2nd to 1st and ground state

So When e go from 2nd(initial) to 1st(final) state then 1/W(21) is

=R{1/2^2-1/3^2}

=R{1/4-1/9}

=R{5/36}-----eqn 1

when e go from 2nd(initial) to ground(final) then 1/W(2g) is

=R{1/1^2-1/3^2}

=R{1-1/9}

=R{8/9}----eqn 2

you have to find

W(21):W(2g)

if you divide eqn 2 to eqn 1 then you find above ratio

so put

{1/w(2g)}/{1/w(21)-----if you solve to find ratio of question(w21/w2g)

put the value

R{8/9}/R{5/36}

R will cancel

=(8/9)*(36/5)

=32/5

So ratio is 32:5

I hope it help

If you find any mistake please  honestly tell me