
The value of Planck's constant has been given as $6.63\times {{10}^{-34}}Js$. The speed of light is also mentioned as $3\times {{10}^{17}}nm{{s}^{-1}}$. What will be the value of wavelength of quantum of light which is closest to the frequency of $6\times {{10}^{15}}{{s}^{-1}}$?
$\begin{align}
& A.10 \\
& B.25 \\
& C.50 \\
& D.75 \\
\end{align}$
Answer
490.2k+ views
Hint: The frequency of the quantum of light is mentioned as the ratio of the velocity of the light to the wavelength of the quantum of light used. Substitute the values given in the question and rearrange the equation in terms of wavelength of light. This will help you answer this question.
Complete step by step answer:
It has been mentioned in the question that the Planck’s constant is given as
$h=6.63\times {{10}^{-34}}Js$
Velocity of light is mentioned in Nanometer. So we have to convert it into metres.
$c=3\times {{10}^{17}}\times {{10}^{-9}}m{{s}^{-1}}$
The frequency has been mentioned as,
$n=6\times {{10}^{-5}}{{s}^{-1}}$
The general equation of the frequency has been given as,
$n=\dfrac{c}{\lambda }$
Where $\lambda $ be the wavelength of the light used. Therefore the wavelength can be found by the equation which has been rearranged as,
$\lambda =\dfrac{c}{n}$
Substituting the values in it will be written as,
$\lambda =\dfrac{3\times {{10}^{17}}\times {{10}^{-9}}}{6\times {{10}^{-5}}}$
Simplifying the equation will give,
$\lambda =5\times {{10}^{-8}}\times {{10}^{9}}nm=50nm$
So, the correct answer is “Option C”.
Note: Frequency is defined as the number of waves falling or incident at a surface in a unit period of time. It is the reciprocal of the time period of the wave. Planck’s constant is a constant in quantum physics which is relating the frequency of a photon and the energy of a photon. The product of the value of Planck’s constant and the frequency of the light incident is equivalent to the energy of the photon used. It is a constant which is named after the famous physicist Max Planck. It is a very important quantity in the context of quantum physics.
Complete step by step answer:
It has been mentioned in the question that the Planck’s constant is given as
$h=6.63\times {{10}^{-34}}Js$
Velocity of light is mentioned in Nanometer. So we have to convert it into metres.
$c=3\times {{10}^{17}}\times {{10}^{-9}}m{{s}^{-1}}$
The frequency has been mentioned as,
$n=6\times {{10}^{-5}}{{s}^{-1}}$
The general equation of the frequency has been given as,
$n=\dfrac{c}{\lambda }$
Where $\lambda $ be the wavelength of the light used. Therefore the wavelength can be found by the equation which has been rearranged as,
$\lambda =\dfrac{c}{n}$
Substituting the values in it will be written as,
$\lambda =\dfrac{3\times {{10}^{17}}\times {{10}^{-9}}}{6\times {{10}^{-5}}}$
Simplifying the equation will give,
$\lambda =5\times {{10}^{-8}}\times {{10}^{9}}nm=50nm$
So, the correct answer is “Option C”.
Note: Frequency is defined as the number of waves falling or incident at a surface in a unit period of time. It is the reciprocal of the time period of the wave. Planck’s constant is a constant in quantum physics which is relating the frequency of a photon and the energy of a photon. The product of the value of Planck’s constant and the frequency of the light incident is equivalent to the energy of the photon used. It is a constant which is named after the famous physicist Max Planck. It is a very important quantity in the context of quantum physics.
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