Question

A certain color of light towards the purple end of the visible spectrum has a wavelength of $420{\text{ nm}}$ in vacuum.What is the frequency of this light? (The speed of light in a vacuum is $3.00 \times {10^8}{\text{m/s}}$ )
$
A.\,{\text{ 126 Hz}} \\
B.\,{\text{ 1}}{\text{.26}} \times {\text{1}}{{\text{0}}^{11}}{\text{Hz}} \\
C.\,{\text{ }}7.1 \times {10^{14}}{\text{ Hz}} \\
D.\,{\text{ 1}}{\text{.4}} \times {\text{1}}{{\text{0}}^{15}}{\text{Hz}} \\
E.\,\;{\text{1}}{\text{.4}} \times {\text{1}}{{\text{0}}^6}{\text{ Hz}} \\
$

Answer 1

Hint: First convert the given unit of the wavelength in the MKS system. Convert nano-metre in metre. Frequency of the light is the rate at which one cycle of repeating vibration occurs in one second. Frequency is measured in Hertz (Hz). One hertz is one vibration cycle per second. Generally, the frequency of the visible light is referred to as the color.

Complete step by step answer:
Given that- the Wavelength of the light is $\lambda = 420{\text{ nm}}$
Convert nano-metres into the metres.
$\therefore \lambda = 420 \times {10^{ - 9}}m$
Speed of the light is $v = 3.00 \times {10^8}{\text{m/s}}$
Now, according to the formula –
$v = \lambda f$
Make the frequency of the light, $f$ as the subject.
$\Rightarrow f = \dfrac{v}{\lambda }$
Place the known values in the above equations –
$\therefore f = \dfrac{{3 \times {{10}^8}}}{{420 \times {{10}^{ - 9}}}}$
When there is the negative power in the denominator, it becomes positive when moved to the numerator.
$
\Rightarrow f = \dfrac{{3 \times {{10}^{8 + 9}}}}{{420}} \\
\Rightarrow f = \dfrac{{3 \times {{10}^{17}}}}{{420}} \\
 $
For easy simplification - Split the power of ten, $17 = 14 + 3$
${10^{17}} = {10^{14}} \times {10^3}$
\[\Rightarrow f = \dfrac{{3 \times 1000 \times {{10}^{14}}}}{{420}} \\
\Rightarrow f = \dfrac{{3000 \times {{10}^{14}}}}{{420}} \\
\Rightarrow f = 7.14 \times {10^{14}}Hz \\
 \]
Therefore, the required solution is - A certain color of light towards the purple end of the visible spectrum has a wavelength of $420{\text{ nm}}$ in vacuum. Then the frequency of the light is \[f = 7.14 \times {10^{14}}Hz\]
Hence, from the given multiple choices – the option C is the correct answer.

Note:: Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.