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ThewavelengthH_a lineis6563Å. Calculatethe wavelengthofFirstmemberof Lymanseries

Answer»

SOLUTION :`(1 )/( lamda _a) = R_n [(1)/(2^2) -(1)/(3^2)]=R_B[1/4 -1/9 ] = (5R_H)/(36 )`
` thereforelamda_a = ( 36 )/( 5R_H) `…….(1)
(a)For LYMAN series`therefore(1)/( lamda_(IL ) ) = R_H[1/(I^2)-(1)/(2^2) ]= (3R_H )/(4 )`
` lamda _(IL ) =(4) /(3R_H)xx (5R_H)/(36 ) = (5)/(27 )`
` lamda _(IL ) =(5 )/(27 )xx lamda_a= (5)/( 27 ) xx 6563= 1215 Å`
(b )forpachenseries`, (1) /(lamda_(IP)) =R_H [(1)/(3^2)-(1)/(4^2)]= (7R_H)/(16xx 9)`
` lamda_(I P) = ( 16xx 9)/(7 R_H)...... (3)`
` therefore(lamda _(IP) )/(lamda_a) = ( 9xx 16 )/(7 R_H) xx (5R_H)/( 36 ) = (20)/(7)`
` lamda _(IP)= (20 )/(7)xx lamda_a( 20xx6563)/(7 ) = 18751Å`


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