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Question

The first member of the Balmer series of hydrogen atom has wavelength of $$656.3\ nm$$, Calculate the wavelength and frequency of the second member of the same series. Given : $$C = 3\times 10^{8}ms^{"1}$$.


Solution

We have
$$\dfrac {1}{\lambda_{0}} = R\left [\dfrac {1}{n_{1}^{2}} - \dfrac {1}{n_{2}^{2}}\right ]$$
For first member of Balmer series
$$n_{1} = 2$$ and $$n_{2} = 3$$
$$\lambda_{1}  = 6563A^{\circ}$$
$$\dfrac {1}{6563\times 10^{-10}} = R\left (\dfrac {1}{2^{2}} - \dfrac {1}{3^{2}}\right ) = \dfrac {5R}{36} ....(1)$$
For second member of Balmer series, $$n_{1} = 2$$ and $$n_{2} = 4$$
$$\dfrac {1}{\lambda_{2}} = R\left (\dfrac {1}{4} - \dfrac {1}{16}\right ) = \dfrac {3R}{16} ....(2)$$
Dividing equation $$(1)$$ by equation $$(2)$$
$$\dfrac {\lambda_{2}}{6563\times 10^{-10}} = \dfrac {5}{36}\times \dfrac {16}{3}$$
$$\lambda_{2} = \dfrac {5\times 16\times 6563\times 10^{-10}}{108} = 4861\ A^{\circ}$$
Frequence $$v = \dfrac {C}{\lambda_{2}} = \dfrac {3\times 10^{8}}{4861\times 10^{-10}} = 0.0006171\times 10^{18}$$
$$v = 6.17\times 10^{14}Hz$$.

Physics

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