
The value of Planck’s constant is \[6.63 \times {10^{ - 34}}\,{\text{J}} - {\text{s}}\]. The velocity of light is \[3 \times {10^8}\,{\text{m}}{{\text{s}}^{ - 1}}\]. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of \[8 \times {10^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}\]?
\[5 \times {10^{ - 18}}\]
\[4 \times {10^1}\]
\[3 \times {10^7}\]
\[2 \times {10^{ - 25}}\]
Answer
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Hint: Use the expression for speed of light in terms of wavelength and frequency of the wave. Substitute the given values of frequency and speed of light and determine the wavelength of a quantum of light.
Formula used:
We know that the speed of light is expressed as,
\[c = \nu \lambda \]
Here, c is the speed of light, \[\lambda \] is the wavelength and \[\nu \] is the frequency.
Complete step by step solution:
We have given the value of Planck’s constant \[h = 6.63 \times {10^{ - 34}}\,{\text{J}} - {\text{s}}\], velocity of light\[c = 3 \times {10^8}\,{\text{m}}{{\text{s}}^{ - 1}}\] and frequency of a quantum of light \[\nu = 8 \times {10^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}\].
We know that the speed of light is expressed as,
\[c = \nu \lambda \]
\[ \Rightarrow \lambda = \dfrac{c}{\nu }\]
Here, c is the speed of light, \[\lambda \] is the wavelength and \[\nu \] is the frequency.
Substituting \[c = 3 \times {10^8}\,{\text{m}}{{\text{s}}^{ - 1}}\] and \[\nu = 8 \times {10^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}\] in the above equation, we get,
\[\lambda = \dfrac{{3 \times {{10}^8}\,{\text{m}}\,{\text{se}}{{\text{c}}^{ - 1}}}}{{8 \times {{10}^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}}}\]
\[ \Rightarrow \lambda = \left( {3.75 \times {{10}^{ - 8}}\,{\text{m}}} \right)\left( {\dfrac{{1\,{\text{nm}}}}{{{{10}^{ - 9}}\,{\text{m}}}}} \right)\]
\[ \Rightarrow \lambda = 3.75 \times {10^1}\,{\text{nm}}\]
\[ \Rightarrow \lambda \approx 4 \times {10^1}\,{\text{nm}}\]
Therefore, the value \[4 \times {10^1}\,{\text{nm}}\] is the closest value of wavelength in nanometers.
So, the correct answer is “Option B”.
Additional Information:
We know that light waves are electromagnetic waves. Electromagnetic waves have frequency, wavelength and velocity since it is a transverse wave. Frequency of the transverse wave is defined as the number of oscillations performed by the wave is one second. Also, the wavelength of the wave is the distance between two consecutive crests of the wave in meters. We know the amplitude of the wave is the maximum vibrations from the mean position.
Note:
Speed of light is constant in vacuum irrespective of frequency and wavelength since the vacuum is not a dispersive medium.
Frequency of light depends on the body that emits light while wavelength of light depends on the medium through which the light is propagating.
Formula used:
We know that the speed of light is expressed as,
\[c = \nu \lambda \]
Here, c is the speed of light, \[\lambda \] is the wavelength and \[\nu \] is the frequency.
Complete step by step solution:
We have given the value of Planck’s constant \[h = 6.63 \times {10^{ - 34}}\,{\text{J}} - {\text{s}}\], velocity of light\[c = 3 \times {10^8}\,{\text{m}}{{\text{s}}^{ - 1}}\] and frequency of a quantum of light \[\nu = 8 \times {10^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}\].
We know that the speed of light is expressed as,
\[c = \nu \lambda \]
\[ \Rightarrow \lambda = \dfrac{c}{\nu }\]
Here, c is the speed of light, \[\lambda \] is the wavelength and \[\nu \] is the frequency.
Substituting \[c = 3 \times {10^8}\,{\text{m}}{{\text{s}}^{ - 1}}\] and \[\nu = 8 \times {10^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}\] in the above equation, we get,
\[\lambda = \dfrac{{3 \times {{10}^8}\,{\text{m}}\,{\text{se}}{{\text{c}}^{ - 1}}}}{{8 \times {{10}^{15}}\,{\text{se}}{{\text{c}}^{ - 1}}}}\]
\[ \Rightarrow \lambda = \left( {3.75 \times {{10}^{ - 8}}\,{\text{m}}} \right)\left( {\dfrac{{1\,{\text{nm}}}}{{{{10}^{ - 9}}\,{\text{m}}}}} \right)\]
\[ \Rightarrow \lambda = 3.75 \times {10^1}\,{\text{nm}}\]
\[ \Rightarrow \lambda \approx 4 \times {10^1}\,{\text{nm}}\]
Therefore, the value \[4 \times {10^1}\,{\text{nm}}\] is the closest value of wavelength in nanometers.
So, the correct answer is “Option B”.
Additional Information:
We know that light waves are electromagnetic waves. Electromagnetic waves have frequency, wavelength and velocity since it is a transverse wave. Frequency of the transverse wave is defined as the number of oscillations performed by the wave is one second. Also, the wavelength of the wave is the distance between two consecutive crests of the wave in meters. We know the amplitude of the wave is the maximum vibrations from the mean position.
Note:
Speed of light is constant in vacuum irrespective of frequency and wavelength since the vacuum is not a dispersive medium.
Frequency of light depends on the body that emits light while wavelength of light depends on the medium through which the light is propagating.
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