Question

Calculate lamnda for extreme line of balmer series of H-spectrum in term of RH

Answer 1

For 1st line in Balmer series n1 =2,n2=3
1/λ1= 109678[ 1/n12 -1/n22  ]
=109678[1/22-1/32]
=109678*5/36 cm-1
=15233 cm-1
λ1=6.56*10-5 cm
wavelength of last line  λ2
1/ λ2=109678[ 1/n12 -1/n22  ]
=109678[ 1/22 -1/∞ ]
λ2=1/109678
=3.65*10-5cm