Question

Find out the wavelength of violet light which has a frequency of $7.5\times {{10}^{14}}Hz$?
$\begin{align}
  & \text{A}\text{. 2}\text{.25}\times \text{1}{{\text{0}}^{23}}m \\
 & \text{B}\text{. 2}\text{.5}\times \text{1}{{\text{0}}^{5}}m \\
 & \text{C}\text{. 4}\times \text{1}{{\text{0}}^{-7}}m \\
 & \text{D}\text{. 7}\text{.5}\times {{10}^{-7}}m \\
 & \text{E}\text{. 2}\text{.25}\times \text{1}{{\text{0}}^{7}}m \\
\end{align}$

Answer 1

Hint: The wavelength and the frequency of any radiation is inversely related. More the frequency less the wavelength. Also wavelength and energy is inversely related. Lower wavelength corresponds to maximum energy as well as maximum frequency.

Formula used:
 Energy of an object having frequency $v$ and wavelength $\lambda $ is
$E=hv=\dfrac{hc}{\lambda }$
Where $h$ is the Planck’s constant and $c$ is the velocity of light.
Their values are $h=6.627\times {{10}^{-34}}JH{{z}^{-1}}$ and $c=3\times {{10}^{8}}m{{s}^{-1}}$
Einstein’s mass energy relation $E=m{{c}^{2}}$

Complete answer:
A light can behave as a particle as wells wave. If we consider light as a wave then like any other wave it must have some wavelength frequency. The wavelength of any wave is the distance between two successive crest or troughs measured in the direction of travel of the wave.
Frequency means the number of times a wave passes through a given cross-sectional area. So the frequency and wavelength are inversely proportional.

According to Planck’s theory light behaves as a particle. The light particles are called photons. The energy associated with each photon with frequency $v$ is given by,

$E=hv=\dfrac{hc}{\lambda }$
Because $v=\dfrac{c}{\lambda }$.

Given the frequency of violet light is $7.5\times {{10}^{14}}Hz$.
So $v=7.5\times {{10}^{14}}Hz$ now wavelength is given by.
$\lambda =\dfrac{c}{v}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{7.5\times {{10}^{14}}Hz}=0.4\times {{10}^{-6}}m=4\times {{10}^{-7}}m$

So the correct option is $\text{C}\text{. 4}\times \text{1}{{\text{0}}^{-7}}m$.

Additional Information:
De-Broglie wavelength: De-Broglie related Planck’s equation with Einstein equation and got a very interesting result.
According to Planck’s theory the energy associated with each photon of frequency $v$ is
$E=hv=\dfrac{hc}{\lambda }$.
According to Einstein the mass energy equivalence relation is given by

$E=m{{c}^{2}}$
So
 $\begin{align}
  & \dfrac{hc}{\lambda }=m{{c}^{2}} \\
 & \Rightarrow \dfrac{h}{\lambda }=mc \\
\end{align}$

If the particle velocity is $V$
Then $\dfrac{h}{\lambda }=mV=p$
P is the momentum.
So the de Broglie wavelength is given by
$\lambda =\dfrac{h}{p}$

Note:
The violet light is a part of visible light. The visible light is a mixture of seven colours .the seven colours are violet, indigo, blue, green, yellow, orange and red. The violet has the smallest wavelength between all visible colours. And red has a maximum wavelength. Because of this, red lights are used as a danger signal because they can travel more distance than any other visible colour.