Answer :
The wavelength of the first line in the Lyman series of the hydrogen spectrum is equal to 121.6 nm and it will be ultraviolet radiation.
When an electron transitions from its lowest energy state of n = 1 to n = 2 (where n is the main quantum number), the Lyman series, a hydrogen spectral series of transitions and ultraviolet emission lines are produced. The transitions are given Greek letter names, such as Lyman-alpha, which goes from n = 2 to n = 1.
Given Z = atomic number and R = Rydberg's constant (109677 cm-1)
since for the first line of the Lyman series, the electron is de-exited from the excited state (i.e., n2=2) to the ground state (i.e., n1=1).
1/∧ =R(1/n1^2- 1/n2^2)
1/∧= 109677(1/1 - 1/4)= 109677(3/4) = 121.6nm
As a result, we know that the wavelength of the first line in the Lyman series of the hydrogen spectrum is equal to 121.6 nm. As the range of ultraviolet radiation is from 100 nm to 400 nm. So it will be ultraviolet radiation.
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