Question

Yellow light emitted from sodium vapour lamps has wavelength 580 nm. Calculate wavelength, frequency and time period of radiation.

Answer 1

Hint: Sodium comes under alkali metals whose ionization energy is very low. So, it tends to release its electron easily. Due to that electron, the radiations are being emitted. By sodium metal, light of yellow colour is emitted. The radiation of light emitted is an electromagnetic wave. The formula to be used here is $\text{c = }\lambda \times \nu $, where c is the speed of light, $\nu $ is the frequency and $\lambda $ is the wavelength.

Complete answer:
-The time taken by periodic motion to complete one cycle is the time period. The S.I. unit is second.
-Frequency is the number of waves completed in a given length of time. It is measured as the number of wave cycles per second or hertz.
- Wavelength is the distance between the two adjacent crests or troughs of a wave. The distance is in meters. Electromagnetic waves or radiations travel with the speed of light. Wavelength, frequency and time period all three are related terms. Let us see their relation:

-Time period is reciprocal of frequency. $\text{Time period = }\dfrac{1}{\text{frequency}}$.
- When wavelength is multiplied to frequency, it gives velocity. $\text{v = f}\times \lambda $, here v is equal to c. So, $\text{c = f}\times \lambda $.

Now, by applying these relations, let us find the answers. Wavelength of radiation is given as 580nm. Frequency will be equal to $\dfrac{3\times {{10}^{8}}}{580\times {{10}^{-9}}}$; nm is ${{10}^{-9}}\text{m}$. The value of frequency is $\text{5}\text{.17}\times \text{1}{{\text{0}}^{14}}\text{ se}{{\text{c}}^{-1}}$. The time period will be $\dfrac{1}{\text{frequency}}$, which is $\dfrac{1}{5.17\times {{10}^{14}}}$ equals $1.93\times {{10}^{-15}}\text{ seconds}$.

The frequency and time period of the radiation is $\text{5}\text{.17}\times \text{1}{{\text{0}}^{14}}\text{ se}{{\text{c}}^{-1}}$ and $1.93\times {{10}^{-15}}\text{ seconds}$.

Note: The value of velocity is taken as $3\times {{10}^{8}}$m/s which is the velocity of light is taken because here we are talking electromagnetic radiation. All the electromagnetic waves travel with the speed of light. Light or any radiation is a type of electromagnetic wave. So, use$\text{v = c = 3}\times \text{1}{{\text{0}}^{8}}\text{m/s}$, when the wave seems to be electromagnetic in nature. Otherwise, $\text{vc}$.