Question 409: Atoms & Nuclei

Question

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:

Accepted Answer

$${1 \over {{\lambda _1}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight)$$ = $${5 \over {36}}$$RZ
2
$${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} ight)$$ = $${{12} \over {64}}$$RZ
2
$$ 	herefore $$ $${{{\lambda _2}} \over {{\lambda _1}}}$$ = $${5 \over {36}} 	imes {{64} \over {12}}$$ = $${{20} \over {27}}$$
$$ \Rightarrow $$ $$\lambda $$
2
= $${{20} \over {27}}{\lambda _1}$$
= $${{20} \over {27}}$$ $$ 	imes $$ 6561 = 4860 $$\mathop A\limits^o $$ = 486 nm

Atoms & Nuclei

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:

Choose the correct answer: