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Grade: 11
        The ratio of difference in wavelength of first and second line of Lyman series in H like atom to difference in wavelength for second and third line of same series is:
2 years ago

Answers : (1)

jitender
114 Points
							1/ ¥ = RZ^2 ( [1/(n1^2) ] - [1/(n2^2)] )¥= wavelengthR = rydburg constant = 109677/cmZ= atomic no.n1= series no.n2= spectral no. From which e is coming.1st line = n1+1 = n22nd line = n1+2 = n23rd line = n1+3 = n3Now For Lyman series n1 = 1Then¥L1= 1/R [1/(1/1^2 - 1/2^2)]=4/3R¥L2= 1/R [1/(1/1^2 - 1/3^2)]=9/8R¥L3= 1/R [1/(1/1^2 - 1/4^2)]=16/15R(¥L1- ¥L2)/(¥L2-¥L3)= [(1/R)[4/3 - 9/8]]/[(1/R)[9/8 - 16/15]] = 25/7
						
2 years ago
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