
Yellow light emitted from a sodium lamp has a wavelength (a) of $580nm$ . Calculate the frequency (v) and wave number (V) of the yellow light.
Answer
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Hint: We must know that the frequency is the number of occurrences in the given time period. Wave number is the reciprocal of wavelength and is also known as the spatial frequency of the wave.
Complete step by step answer:
Let’s start with discussing the different parameters of light that we need to find. Frequency is the number of occurrences in the given time period. Wave number is the reciprocal of wavelength and is also known as the spatial frequency of the wave.
Now we are provided with the information that we are having a sodium lamp which radiates yellow light of wavelength 580 nm. We need to find frequency and wave numbers. As we know frequency is the speed of substance per unit wavelength. So, in case of yellow light the frequency will be
Frequency $ = \dfrac{{{\text{Speed of light}}}}{{{\text{Wavelength of yellow light}}}}$
On substituting the known values in the above given formula we get,
$ \Rightarrow {\text{v = }}\dfrac{{3 \times {{10}^8}{\text{ m/s}}}}{{5.8 \times {{10}^{ - 7}}{\text{ m}}}}$
On simplification of the above fraction we get,
$v = 5.17 \times {10^{14}}{s^{ - 1}}$
Hence, the frequency of the yellow light is $5.17 \times {10^{14}}{s^{ - 1}}$ .
We have to remember that the wave number is defined as the reciprocal of wavelength so,
Wavenumber = ${\text{V = }}\dfrac{1}{{{\text{Wavelength of yellow light}}}}$
On substituting the known values we get,
$ \Rightarrow {\text{V = }}\dfrac{1}{{5.8 \times {{10}^{ - 7}}{\text{ m}}}}$
On simplifying we get,
$ \Rightarrow 1.7 \times {10^6}{m^{ - 1}}$
Hence, the wave number of the yellow light is $1.7 \times {10^6}{m^{ - 1}}$ .
Note:
We have to remember that the frequency and wave number along with wavelength, all three are the properties of a wave. The light shows a dual nature which is of both the particle and the wave. Long with these three properties, a wave is having many other properties such as Reflection, refraction, absorption, etc.
Complete step by step answer:
Let’s start with discussing the different parameters of light that we need to find. Frequency is the number of occurrences in the given time period. Wave number is the reciprocal of wavelength and is also known as the spatial frequency of the wave.
Now we are provided with the information that we are having a sodium lamp which radiates yellow light of wavelength 580 nm. We need to find frequency and wave numbers. As we know frequency is the speed of substance per unit wavelength. So, in case of yellow light the frequency will be
Frequency $ = \dfrac{{{\text{Speed of light}}}}{{{\text{Wavelength of yellow light}}}}$
On substituting the known values in the above given formula we get,
$ \Rightarrow {\text{v = }}\dfrac{{3 \times {{10}^8}{\text{ m/s}}}}{{5.8 \times {{10}^{ - 7}}{\text{ m}}}}$
On simplification of the above fraction we get,
$v = 5.17 \times {10^{14}}{s^{ - 1}}$
Hence, the frequency of the yellow light is $5.17 \times {10^{14}}{s^{ - 1}}$ .
We have to remember that the wave number is defined as the reciprocal of wavelength so,
Wavenumber = ${\text{V = }}\dfrac{1}{{{\text{Wavelength of yellow light}}}}$
On substituting the known values we get,
$ \Rightarrow {\text{V = }}\dfrac{1}{{5.8 \times {{10}^{ - 7}}{\text{ m}}}}$
On simplifying we get,
$ \Rightarrow 1.7 \times {10^6}{m^{ - 1}}$
Hence, the wave number of the yellow light is $1.7 \times {10^6}{m^{ - 1}}$ .
Note:
We have to remember that the frequency and wave number along with wavelength, all three are the properties of a wave. The light shows a dual nature which is of both the particle and the wave. Long with these three properties, a wave is having many other properties such as Reflection, refraction, absorption, etc.
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