Question

The value of Planck’s constant is $6.63\times {{10}^{-34}}$ Js. 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 $8\times {{10}^{15}}{{\sec }^{-1}}$ ?
(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 1

Hint: There is a relation between the velocity of the light, the wavelength of the light with the frequency of the light. The relationship between velocity, wavelength and frequency is as follows.
As per relation
$v=\dfrac{c}{\lambda }$
Here $v$ = frequency of the light
$c$ = velocity of the light.
$\lambda $ = wavelength of the light.

Complete step by step solution:
-In the question it is given that velocity of light is $3\times {{10}^{8}}$ m/sec (c) and frequency of light is$8\times {{10}^{15}}{{\sec }^{-1}}$ ($v$).
-With the given data we have to find the closest to the wavelength of a quantum of light.
-The formula to calculate the closest to the wavelength of a quantum of light is as follows.
$\begin{align}
& v=\dfrac{c}{\lambda } \\
 & \lambda =\dfrac{c}{v} \\
\end{align}$
-Substitute the given values in the above equation to get the wavelength of a quantum of light.
$\begin{align}
& \lambda =\dfrac{c}{v} \\
& \lambda =\dfrac{3\times {{10}^{8}}}{8\times {{10}^{15}}} \\
& \lambda =3.754\times {{10}^{-8}}m\approx 4.0\times {{10}^{-8}}m \\
\end{align}$
-Therefore the closest or nearest to the wavelength of a quantum of light is$4.0\times {{10}^{-8}}m$.

So, the correct option is (B).

Note: Wavelength of the light is inversely proportional to the frequency of the light and directly proportional to the velocity of the light. As the frequency of the light increases, the wavelength of the light decreases and the velocity of the light increases.
The unit for velocity is m/sec.
The unit for wavelength is nm or m.
The unit for frequency is${{\sec }^{-1}}$.