JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 22)
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
486
Explanation
$${1 \over {{\lambda _1}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$$ = $${5 \over {36}}$$RZ2
$${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$ = $${{12} \over {64}}$$RZ2
$$ \therefore $$ $${{{\lambda _2}} \over {{\lambda _1}}}$$ = $${5 \over {36}} \times {{64} \over {12}}$$ = $${{20} \over {27}}$$
$$ \Rightarrow $$ $$\lambda $$2 = $${{20} \over {27}}{\lambda _1}$$
= $${{20} \over {27}}$$ $$ \times $$ 6561 = 4860 $$\mathop A\limits^o $$ = 486 nm
$${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$ = $${{12} \over {64}}$$RZ2
$$ \therefore $$ $${{{\lambda _2}} \over {{\lambda _1}}}$$ = $${5 \over {36}} \times {{64} \over {12}}$$ = $${{20} \over {27}}$$
$$ \Rightarrow $$ $$\lambda $$2 = $${{20} \over {27}}{\lambda _1}$$
= $${{20} \over {27}}$$ $$ \times $$ 6561 = 4860 $$\mathop A\limits^o $$ = 486 nm
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