
The value of Planck's constant is $ 6.63 \times {10^{ - 34}} $ $ {{J - sec}} $ . The velocity of light is $ 3 \times {10^8} $ $ {{m/sec}} $ . Which value is closest to the wavelength of a quantum of light with frequency of $ 8 \times {10^{15}} $ $ {{se}}{{{c}}^{{{ - 1}}}} $ ?
(A) $ 5 \times {10^{ - 18}} $ m
(B) $ 4 \times {10^{ - 8}} $ m
(C) $ 3 \times {10^7} $ m
(D) $ 2 \times {10^{ - 25}} $ m
Answer
444.6k+ views
Hint: This problem is based on the relation between the wavelength of a light radiation with the speed of the light and the frequency of the vibration of the photons of the light radiation, where the photons are the particles that are present in the light radiation.
Formula: $ {{c = }}\nu {{ }}\lambda $
Where c is the speed of light, $ \nu $ is the frequency of light radiation, and $ \lambda $ is the wavelength of the light radiation.
Complete stepwise Solution
We know that the speed of any object is equal to the distance covered by time taken. The frequency is defined as the number of times any event happens within one second or one minute or one hour, so the frequency is inversely proportional to the time period. The wavelength of a radiation or a light wave is defined as the distance travelled to complete a full wave or in case of light it is defined as the distance between two crests or two troughs. The speed of light is related to the wavelength and the frequency of the light radiation according to the following equation:
$ {{c = }}\nu {{ }}\lambda $
Rearranging the equation, we have,
$ \lambda {{ = }}\dfrac{{{c}}}{{{\nu }}} $
Putting the values of the speed of light and the frequency of the radiation we have,
$ \lambda = \dfrac{{3 \times {{10}^8}}}{{8 \times {{10}^{15}}}} = 3.75 \times {10^{ - 8}} $ m
This value of the wavelength if rounded off to the closest number is: $ 4 \times {10^{ - 8}} $ m
Hence, the correct answer is option B.
Note
The particle nature of the light radiation is a consequence of de-Broglie's equation and is based on the fact that matter possesses dual nature of particle as well as wave and likewise light also has dual nature of both particle and wave. Mathematically the de-Broglie equation is expressed as,
$ \lambda {{ = }}\dfrac{{{h}}}{{{{mv}}}} $
Where $ {{\lambda }} $ is the wavelength of the light, v is the velocity of light, h is Planck's constant.
Formula: $ {{c = }}\nu {{ }}\lambda $
Where c is the speed of light, $ \nu $ is the frequency of light radiation, and $ \lambda $ is the wavelength of the light radiation.
Complete stepwise Solution
We know that the speed of any object is equal to the distance covered by time taken. The frequency is defined as the number of times any event happens within one second or one minute or one hour, so the frequency is inversely proportional to the time period. The wavelength of a radiation or a light wave is defined as the distance travelled to complete a full wave or in case of light it is defined as the distance between two crests or two troughs. The speed of light is related to the wavelength and the frequency of the light radiation according to the following equation:
$ {{c = }}\nu {{ }}\lambda $
Rearranging the equation, we have,
$ \lambda {{ = }}\dfrac{{{c}}}{{{\nu }}} $
Putting the values of the speed of light and the frequency of the radiation we have,
$ \lambda = \dfrac{{3 \times {{10}^8}}}{{8 \times {{10}^{15}}}} = 3.75 \times {10^{ - 8}} $ m
This value of the wavelength if rounded off to the closest number is: $ 4 \times {10^{ - 8}} $ m
Hence, the correct answer is option B.
Note
The particle nature of the light radiation is a consequence of de-Broglie's equation and is based on the fact that matter possesses dual nature of particle as well as wave and likewise light also has dual nature of both particle and wave. Mathematically the de-Broglie equation is expressed as,
$ \lambda {{ = }}\dfrac{{{h}}}{{{{mv}}}} $
Where $ {{\lambda }} $ is the wavelength of the light, v is the velocity of light, h is Planck's constant.
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