First slide

Bohr Model of the Hydrogen atom.

Question

If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 $\overset{o}{A}$ , the wavelength of second member of Balmer series will be :

Easy

A

1215 $\overset{o}{A}$
B

4862 $\overset{o}{A}$
C

6050 $\overset{o}{A}$
D

data given is insufficient to calculate the value

Solution

$\frac{1}{λ} α (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{\frac{1}{2^{2}} - \frac{1}{4^{2}}}{\frac{1}{2^{2}} - \frac{1}{3^{2}}} \Rightarrow \frac{6564}{λ_{2}} = \frac{27}{20}; ∴ λ_{2} = 4862 A^{\circ}$

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