The wavelengths in the hydrogen spectrum with m=1form a series of spectral lines called the Lyman series. Calculate the wavelengths of the first four members of the Lyman series.

Short Answer

Expert verified

The wavelengths of the first four members of the Lyman series are :

For n=2; λ=121.57nm

For n=3; λ=102.57nm

For n=4; λ=97.26nm

Forn=5;λ=94.98nm

Step by step solution

01

Given information

We need to find the wavelengths of the first four members of the Lyman series.

02

Simplify

The expression for the Lyman series is:

λ=91.158(nm)1-1n2, where (n=2,3,4,...) (Let this equation be (1) )

Here, λis the wavelength of the electron emitted and nis the principle quantum number of the shell.

For n=2:

λ=91.158(nm)1-1(2)2 (By substituting n=2in equation (1))

λ=121.57nm

For n=3:

λ=91.158(nm)1-1(3)2 (By substituting n=3in equation (1))

λ=102.57nm

For n=4:

λ=91.158(nm)1-1(4)2 (By substituting n=4in equation (1))

λ=97.26nm

For n=5:

λ=91.158(nm)1-1(5)2 (By substituting n=5in equation (1))

λ=94.98nm

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