The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915\mathop A\limits^o$. The longest wavelength of spectral lines in the Balmer series will be _______ $\mathop A\limits^o$.
Answer (integer)
6588
Solution
<p>$\frac{1}{915}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\quad$ (For Lyman)</p>
<p>$\Rightarrow \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\quad$ (For Balmer)</p>
<p>$\Rightarrow \lambda=6588$ $\mathop A\limits^o$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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