Calculate energy,frequency and wavelength of radiation which is corresponding to the spectral line of lowest frequency in lyman series in hydrogen atom spectrum
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Answer:
121.6nm
Explanation:
1λ=R(1(n1)2−1(n2)2)⋅Z2
where,
R = Rydbergs constant (Also written is RH)
Z = atomic number
Since the question is asking for 1st line of Lyman series therefore
n1=1
n2=2
since the electron is de-exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series.

Therefore plugging in the values
1λ=R(1(1)2−1(2)2)⋅12
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
λ=43⋅912.A
since 1R=912.A
therefore
λ=1216.A
or
λ=121.6nm
121.6nm
Explanation:
1λ=R(1(n1)2−1(n2)2)⋅Z2
where,
R = Rydbergs constant (Also written is RH)
Z = atomic number
Since the question is asking for 1st line of Lyman series therefore
n1=1
n2=2
since the electron is de-exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series.

Therefore plugging in the values
1λ=R(1(1)2−1(2)2)⋅12
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
λ=43⋅912.A
since 1R=912.A
therefore
λ=1216.A
or
λ=121.6nm
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