Question

Compare the energies of two radiations ${E_1}$ with wavelength $800{\text{ nm}}$ and ${E_2}$ with wavelength $400{\text{ nm}}$.
A.${{\text{E}}_1}{\text{ = 2}}{{\text{E}}_2}$
B. ${{\text{E}}_1}{\text{ = }}{{\text{E}}_2}$
C. ${{\text{E}}_2}{\text{ = 2}}{{\text{E}}_1}$
D. ${{\text{E}}_2}{\text{ = - }}\dfrac{1}{2}{{\text{E}}_1}$

Answer 1

When electromagnetic waves travel, then they carry some amount of energy with them. The energy carried by the electromagnetic waves is given by using Einstein’s equation. According to which the energy of electromagnetic radiation is the product of Planck’ constant and the frequency of the waves. We will compare the energies by using this relation.

Formula Used:
${\text{E = }}hv$
Where, $E$ is the energy possessed by the radiation, $h$ is known as Planck’s constant and $v$ is the frequency of radiation.

Complete step by step answer:
When an electromagnetic wave travels in a medium then it carries an amount of energy with it. The amount of energy carried by the wave is the product of Planck’s constant and the frequency with which it travels in the medium. According to Einstein equation,
$\Rightarrow {\text{ E = }}hv$
Here $v$ is the frequency of radiation and we know that frequency of wave is the ratio of speed of light and distance travelled by it in one cycle which is known as wavelength of wave.

Therefore it can be written as,
$\Rightarrow {\text{ }}v{\text{ = }}\dfrac{c}{\lambda }$
Therefore it can be written as,
$\Rightarrow {\text{ E = }}h\dfrac{c}{\lambda }$
From above we can observe that the value of speed of light and Planck’s constant are always constant. Hence we can say that energy carried by electromagnetic radiation is inversely proportional to its wavelength.
$\Rightarrow {\text{ E = }}\dfrac{1}{\lambda }$

According to question Energy for wavelength $800{\text{ nm}}$ can be calculated as:
$\Rightarrow {\text{ }}{{\text{E}}_1}{\text{ = }}\dfrac{1}{\lambda }$
$\Rightarrow {\text{ }}{{\text{E}}_1}{\text{ = }}\dfrac{1}{{800}}$ ____________$(1)$
Similarly for wavelength $400{\text{ nm}}$ energy will be:
$\Rightarrow {\text{ }}{{\text{E}}_2}{\text{ = }}\dfrac{1}{\lambda }$
$\Rightarrow {\text{ }}{{\text{E}}_2}{\text{ = }}\dfrac{1}{{400}}$ _____________$(2)$
On dividing equation $(1)$ and equation $(2)$ we get the result as:
$\Rightarrow {\text{ }}\dfrac{{{{\text{E}}_1}}}{{{{\text{E}}_2}}}{\text{ = }}\dfrac{{400}}{{800}}$
On solving we get the reduced result as:
$\therefore {\text{ }}\dfrac{{{{\text{E}}_1}}}{{{{\text{E}}_2}}}{\text{ = }}\dfrac{1}{2}$ Or ${{\text{E}}_2}{\text{ = 2}}{{\text{E}}_1}$

Therefore the correct option is C.

Note: We can remember the relation between the energy and wavelength of electromagnetic radiations. Since we have to find the ratio of both energies so do not put the value of speed of light and Planck’s constant. Thus we can say that the wavelength having more wavelength will have lesser energy with it.