omreknaqec7

2023-02-18

Yellow light emitted from a sodium lamp has a wavelength ($\lambda $) of 580 nm. Calculate the frequency (ν) and wave number ($\overline{V}$) of the yellow light.

a4dlity3v

Beginner2023-02-19Added 6 answers

From the wording,

$\lambda =\frac{c}{v}$

we get,$v=\frac{c}{\lambda}$.............(i)

Where, ν = frequency of yellow light

c = velocity of light in vacuum $=3\times {10}^{8}\text{}m/s$

$\lambda $ = wavelength of yellow light $=580nm=580\times {10}^{-9}\text{}m$ Substituting the values in expression (i):

$v=\frac{3\times {10}^{8}}{580\times {10}^{-9}}=5.17\times {10}^{14}{s}^{-1}$

As a result, the sodium lamp's yellow light emission frequency $=5.17\times {10}^{14}{s}^{\u20131}$

Wave number of yellow light $\overline{v}=\frac{1}{\lambda},\phantom{\rule{0ex}{0ex}}=\frac{1}{580\times {10}^{-9}}=1.72\times {10}^{6}{m}^{-1}$

$\lambda =\frac{c}{v}$

we get,$v=\frac{c}{\lambda}$.............(i)

Where, ν = frequency of yellow light

c = velocity of light in vacuum $=3\times {10}^{8}\text{}m/s$

$\lambda $ = wavelength of yellow light $=580nm=580\times {10}^{-9}\text{}m$ Substituting the values in expression (i):

$v=\frac{3\times {10}^{8}}{580\times {10}^{-9}}=5.17\times {10}^{14}{s}^{-1}$

As a result, the sodium lamp's yellow light emission frequency $=5.17\times {10}^{14}{s}^{\u20131}$

Wave number of yellow light $\overline{v}=\frac{1}{\lambda},\phantom{\rule{0ex}{0ex}}=\frac{1}{580\times {10}^{-9}}=1.72\times {10}^{6}{m}^{-1}$