Question
If the wavelength of first member of Balmer series of hydrogen spectrum is 6564˙A, the wavelength of second member of Balmer series will be:
Answer: Option B
:
B
1λ1∝[122−132]
1λ1∝536 . . . .(1)
1λ2∝[14−116]
1λ2∝[316] . . . .(2)
λ2λ1=5×1636×3=2027
λ2=2027×6564=4862˙A
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B
1λ1∝[122−132]
1λ1∝536 . . . .(1)
1λ2∝[14−116]
1λ2∝[316] . . . .(2)
λ2λ1=5×1636×3=2027
λ2=2027×6564=4862˙A
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