Calculate shortest and longest wavelength in hydrogen spectrum in balmer series R=109677.6cm^-1
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Answered by
1
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.
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.
Answered by
0
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.
Explanation:
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