JEE MAIN - Physics (2024 - 5th April Evening Shift - No. 27)
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 _______ $$\mathop A\limits^o$$.
Respondre
6588
Explicació
$$\frac{1}{915}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\quad$$ (For Lyman)
$$\Rightarrow \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\quad$$ (For Balmer)
$$\Rightarrow \lambda=6588$$ $$\mathop A\limits^o $$