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The first line in the Lyman series has wavelength \lambda.The wavelength of the first line in Balmer series is

Option: 1

\frac{2}{9}\lambda


Option: 2

\frac{9}{2}\lambda


Option: 3

\frac{5}{27}\lambda


Option: 4

\frac{27}{5}\lambda


Answers (1)

best_answer

\mathrm{\text{For first line in Lyman series}\, \lambda_{L_1}=\frac{4}{3 R}}\quad \cdots(i)

\mathrm{\text{For first line in Balmer series}\, \lambda_{B_1}=\frac{36}{5 R}}\quad \cdots\left ( ii \right )

From equation (i) and (ii)

\mathrm{\frac{\lambda_{B_1}}{\lambda_{L_1}}=\frac{27}{5} \Rightarrow \lambda_{B_1}=\frac{27}{5} \lambda_{L_1} \Rightarrow \lambda_{B_1}=\frac{27}{5} \lambda}

Posted by

Gautam harsolia

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