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Question-13627




Question Number 13627 by tawa tawa last updated on 21/May/17
Answered by ajfour last updated on 21/May/17
(1/λ)=R((1/n_2 ^2 )−(1/n_1 ^2 ))  (1/λ)=R((1/4)−(1/9))  λ=((36)/(5R))=((36)/(5(1.097×10^7 m^(−1) )))  λ=((3600)/(5×1.097)) nm=((720)/(1.097))nm     λ≈656nm .
$$\frac{\mathrm{1}}{\lambda}={R}\left(\frac{\mathrm{1}}{{n}_{\mathrm{2}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}_{\mathrm{1}} ^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{\lambda}={R}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$\lambda=\frac{\mathrm{36}}{\mathrm{5}{R}}=\frac{\mathrm{36}}{\mathrm{5}\left(\mathrm{1}.\mathrm{097}×\mathrm{10}^{\mathrm{7}} {m}^{−\mathrm{1}} \right)} \\ $$$$\lambda=\frac{\mathrm{3600}}{\mathrm{5}×\mathrm{1}.\mathrm{097}}\:{nm}=\frac{\mathrm{720}}{\mathrm{1}.\mathrm{097}}{nm} \\ $$$$\:\:\:\lambda\approx\mathrm{656}{nm}\:. \\ $$
Commented by tawa tawa last updated on 21/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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