Question

The value of Planck's constant is $6.63\times {{10}^{-34}}Js$. The velocity of light is $3\times {{10}^{8}}m\,{{s}^{-1}}$. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $8\times {{10}^{15}}{{s}^{-1}}$.
A.$4\times {{10}^{1}}$
B.$4\times {{10}^{4}}$
C.$3\times {{10}^{3}}$
D.$2\times {{10}^{2}}$

Answer 1

Hint: Use the Planck's equation to find out the wavelength from the values given in the question and then round off to the nearest figure.

Complete answer:
In order to answer our question, we need to learn about the particle nature of electromagnetic radiation and some of its elements. According to classical mechanics, energy is emitted or absorbed discontinuously. Therefore, the energy of any electromagnetic radiation is proportional to its intensity and independent of its frequency or wavelength. Thus, the radiation emitted by the body being heated should have the same colour throughout, although, the intensity of the colour might change with variation in temperature.
Thereafter Max Planck suggested that the energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy. These packets of energy are named as quanta'. In case of light, the quantum of energy is called a 'photon'. The energy of radiation is proportional to its frequency ($\upsilon $) and is expressed by equation:
\[E=\dfrac{hc}{\lambda }\]
Here, E is the energy emitted by the photoelectron, h is the Planck's constant, c is the velocity of light and $\lambda $is the wavelength. Also, we have $\nu =\dfrac{c}{\lambda }$, using this equation we can find the value of $\lambda $, which is:
\[\lambda =\dfrac{3\times {{10}^{8}}}{8\times {{10}^{15}}}=0.375\times {{10}^{-7}}m\]
On estimating, we get the value of wavelength close to $4\times {{10}^{1}}$.

So, we get option A as the correct answer for the question.

Note:
It is to be noted that the energy which is liberated by the photoelectrons consist of heat, light and kinetic energy. The kinetic energy is given by the formula $K.E=\dfrac{1}{2}m{{v}_{0}}^{2}$.