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## Q. 6.3

Calculate the wavelength in nanometers of the line in the Balmer series that results from the transition n = 4 to n = 2.

 ANALYSIS n = 2; n = 4 Information given: speed of light $(2.998 × 10^{8} m/s)$ Rydberg constant $(2.180 \times 10^{-18} J)$ Planck constant $(6.626 ×10^{-34} J \cdot s)$ Information implied: wavelength in nm Asked for:

STRATEGY

1. Substitute into Equation 6.4 to find the frequency due to the transition.
$\nu =\frac{R_H}{h}\left(\frac{1 }{n^{2}_{lo} }-\frac{1}{n^{2}_{hi}} \right)$

Use the lower value for n as nlo and the higher value for $n_{hi}$ .

2. Use Equation 6.1 to find the wavelength in meters and then convert to nanometers

$\lambda \nu =c$          (6.1)

## Verified Solution

1. Frequency       $\nu =\frac{2.180 \times 10^{-18} J}{6.626 ×10^{-34} J \cdot s}\left(\frac{1 }{(2)^{2} }-\frac{1}{(4)^{2} } \right)= 6.169×10^{14 } s^{-1}$

2. Wavelength          $\lambda =\frac{2.998 \times 10^{8} m/s}{6.169 \times 10^{14} s^{-1} } \times \frac{1 nm}{ 1 \times 10^{-9} m} =486.0 nm$

END POINT

Compare this value with that listed in Table 6.2 for the second line of the Balmer series.