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Chapter 6

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)

Step-by-Step

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.