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Question

Chemistry Question on Structure of atom

What is the wave number of 4th4^{th} line in Balmer series of hydrogen spectrum? (R=1,09,677 cm−1)(R=1,09,677\,cm^{-1})

A

24,630 cm−1 24,630\,cm^{-1}

B

24,360 cm−1 24,360\,cm^{-1}

C

24,730 cm−1 24,730\,cm^{-1}

D

24,372 cm−1 24,372\,cm^{-1}

Answer

24,372 cm−1 24,372\,cm^{-1}

Explanation

Solution

vˉ=R(1n12−1n22)\bar{v}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) For Balmer series, n1=2n_{1}=2 and for 4th 4^{\text {th }} line in Balmer series n2=6n_{2}=6 ∵R=109677 cm−1\because R =109677\, cm ^{-1} vˉ=109677(122−162)\bar{v} =109677\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}\right) =109677(14−136)=109677\left(\frac{1}{4}-\frac{1}{36}\right) vˉ=24,372 cm−1\bar{v} =24,372\, cm ^{-1}