The wave number of the first line in the Balmer series of hydrogen is `15200 cm^(-1)`. What would be the wavenumber of the first line in the Lyman ser

For series limit of Balmer series, we can write, `n_(2)=2,n_(1)=oo` `rArr " " (1)/(lambda)=R((1)/(n_(2)^(2))-(1)/(n_(1)^(2)))=R[(1)/((2)^(2))-(1)/((oo)^(2))]=(R)/(4)` `:. " " lambda=(4)/(R) =(4)/(10967800)=4xx912xx10^(-10)m` `=4xx912Å=3648Å`

Wavelength of the first line of Lyman series for hydrogen atom (Z=1) will be given by the equation, `(1)/(lambda)=R((1)/(1^(2))-(1)/(2^(2)))=(3R)/(4)` The wavelength of second Balmer line for hydrogen like ion Y...

Correct Answer - A `(lambda_("Lyman"))/(lambda_("Blamer"))=(((1)/2^(2)-(1)/(3^(2))))/(((1)/(1^(2))-(1)/(2^(2))))=(5)/(27)" " [:. (1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))]` `rArrlambda_("Lyman")=(5)/(27)xxlambda_("Balmer")=(5)/(27)xx6563=1215.4Å`

Correct Answer - C Here`lambda_L=1215Å` for the first line of lyman series. `(1)/(lambda_L)=R[1/(1^2)-1/(2^2)]=R[1-1/(4)]=(3R)/(4)` `therefore lambda_L=(4)/(3R)` for first line of balmer series `(1)/(lambda_B)=R[1/(2^2)-1/(3^2)]=R[(1)/(4)-1/(9)]rArr(1)/(lambda_B)=R[5/(36)]` `therefore lambda_B=(36)/(5R)` From (i) and (ii), `(lambda_B)/(lambda_L)=(36//5R)/(4//3R)=(36xx3)/(4xx5)` `therefore lambda_B=(108)/(20)xxlambda_L=(108)/(20)xx1215=6561Å`

Correct Answer - A For a balmer series `(1)/(lambda_B)=R[(1)/(2^2)-(1)/(n^2)]`........(i) where n=3,4,.... by putting `n=oo` in equation (i), we obtain the series limit of the balmer series. This is the shortest wavelength...

Correct Answer - A For lyman series `upsilon=Rc[(1)/(1^2)-(1)/(n^2)]` where n=2,3,4,....... for the seires limit of lyman series , `n=oo` `therefore upsilon_1=Rc[(1)/(1^2)-(1)/(oo^2)]=Rc`.........(i) for the first line of lyman series, `n=2` `therefore upsilon_2=Rc[(1)/(1^2)-(1)/(2^2)]=(3)/(4)Rc`........(ii)...

Correct Answer - A If `lambda_D` and `lambda_H` are wavelength emitted in the case of deuterium and hydrogen `therefore (lambda_D)/(lambda_H)=(1-(m_e)/(2m_p))` Here, `lambda_H=1028Å, m_p=1860m_e` `therefore lambda_D=(1-(1)/(2xx1860))lambda_H=(1-(1)/(3720))lambda_H` `lambda _D0.9997lambda_H=0.99973xx1028Å=1027.7Å`.

Correct Answer - A Wave number `1/(lambda) = 1.097 xx 10^(7) [(1)/(1^(2)) - (1)/(n^(2))]` Now in series limit `lambda` corresponds to `n = oo` to `n =1` `:.` Wave number for...

Correct Answer - 1215.14 Å