Question:
Taking the wavelength of first Balmer line in hydrogen spectrum $(\mathrm{n}=3$ to $\mathrm{n}=2)$ as $660 \mathrm{~nm}$, the wavelength of the $2^{\text {nd }}$ Balmer line $(n=4$ to $n=2)$ will be;
Correct Option: , 2
Solution:
(2) $\frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36}$
$\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 R}{16}$
$\therefore \frac{\lambda_{2}}{\lambda_{1}}=\frac{80}{108}$
$\lambda_{2}=\frac{80}{108} \lambda_{1}=\frac{80}{108} \times 660=488.9 \mathrm{~nm}$
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