Taking the wavelength of first Balmer line in hydrogen spectrum
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;

  1. (1) $889.2 \mathrm{~nm}$

  2. (2) $488.9 \mathrm{~nm}$

  3. (3) $642.7 \mathrm{~nm}$

  4. (4) $388.9 \mathrm{~nm}$


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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