Question

# The wavelength of the first line in Balmer series in the hydrogen spectrum is λ. What is the wavelength of the second line?

A
2027λ
B
316λ
C
536λ
D
34λ

Solution

## The correct option is B 2027λThe first line in Balmer series corresponds to n=3. ∴1λ=R(122−132)=R(14−19)=5R36 ⇒λ=365R For the second line, n=4. ∴1λ′=R(122−142)=R(14−116)=3R16 ⇒λ′=163R Now, λ′λ=163R×5R36=2027⇒λ′=2027λ Hence, the correct choice is (a).

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