The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:
Answer (integer)
486
Solution
$${1 \over {{\lambda _1}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$$ = ${5 \over {36}}$RZ<sup>2</sup>
<br><br>$${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$ = ${{12} \over {64}}$RZ<sup>2</sup>
<br><br>$\therefore$ ${{{\lambda _2}} \over {{\lambda _1}}}$ = ${5 \over {36}} \times {{64} \over {12}}$ = ${{20} \over {27}}$
<br><br>$\Rightarrow$ $\lambda$<sub>2</sub> = ${{20} \over {27}}{\lambda _1}$
<br><br>= ${{20} \over {27}}$ $\times$ 6561 = 4860 $\mathop A\limits^o$ = 486 nm
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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