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Given R_H =1.097 xx 10^7 m^(-1)findthe longest and shortest wavelengthlimitof Pashenseriesfor longest wavelengthn_i =3 andn_f = 4forshortestwavelengthn_i =3 andn_f= oo( infinity) |
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Answer» Solution :`vec(UPSILON ) = (1)/( lamda ) = R_H[(1)/(n_(i)^(2))-(1)/(n_(f)^(2))]` ` (1)/(lamda_("long"))=R_H[1/(3^2) - (1)/(4^2)]` ` thereforelamda_("long")=( 16xx 9 )/( 7R_H)=( 16xx 9 )/( 7 XX 1.097 xx 10^7 ) = 10752Å` ` (1 )/( lamda_("short")) = R_H = [1/9 - 1/oo ]= (R_H)/(9)` ` lamda_("short")= (9)/( R_H ) = (9)/( 1.097 xx 10^7 ) = 8204 A` |
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