Question

The energy per quantum associated with the light of wavelength \[250\text{ }\times \text{ }{{10}^{-9\text{ }}}\text{m }\] is:
A) \[\text{ 7}\text{.95 }\times \text{ 10}{{\text{ }}^{-19}}\text{ J }\]
B) \[\text{ 7}\text{.95 }\times \text{ 10}{{\text{ }}^{-26}}\text{ J }\]
C) \[\text{ 3}\text{.93 }\times \text{ 10}{{\text{ }}^{-26}}\text{ J }\]
D) \[\text{ 3}\text{.93 }\times \text{ 10}{{\text{ }}^{-19}}\text{ J }\]

Answer 1

Hint: According to planck's quantum theory, the energy of the electromagnetic radiation (E) is directly related to the frequency of radiation$\text{ }\nu \text{ }$. The planck's constant ‘h’ is the proportionality factor. The value of Planck's constant is equal to the\[\text{ 6}\text{.626 070 15 }\times \text{ 1}{{\text{0}}^{\text{-}}}^{\text{34}}\text{ }\!\!~\!\!\text{ J s}\].

Complete step by step answer:
According to Plank’s quantum theory,
1) The atoms and molecules emit or absorb the energy in the packets. These energy packets are called a quanta. This is discrete quantities of energies.
The smallest packet of energy that can be absorbed or emitted by the electromagnetic radiation is termed as a quantum.
2) The energy which is absorbed or emitted is found to be directly related to the frequency $\text{ }\nu \text{ }$ of the radiation.
Thus, we can derive a relation which relates the energy with the frequency of the radiation as follows,
$\text{ E = h }\nu $ …………………………………... (1)
Where,
E is the energy
The ‘h’ is the Planck's constant .The value of planck's constant is \[\text{ 6}\text{.626 070 15 }\times \text{ 1}{{\text{0}}^{\text{-}}}^{\text{34}}\text{ }\!\!~\!\!\text{ J s}\].
The $\text{ }\nu \text{ }$is the frequency of the radiation.
This is called the planck's equation.
We are given the following data.
The wavelength of the radiation \[\text{ }\!\!\lambda\!\!\text{ }=\text{ }250\text{ }\times \text{ }{{10}^{-9\text{ }}}\text{m }\]
To find the energy of the radiation.
We know that the speed of the light (C) is equal to the product of the frequency $\text{ }\nu \text{ }$and the wavelength \[\text{ }\!\!\lambda\!\!\text{ }\]of the radiation. Mathematically, the relation is stated as follows,
$\text{ C = }\!\!\nu\!\!\text{ }\!\!\times\!\!\text{ }\!\!\lambda\!\!\text{ }$
On rearrangement we have,
$\text{ }\!\!\nu\!\!\text{ = }\dfrac{\text{C}}{\text{ }\!\!\lambda\!\!\text{ }}$ (2)
Substitute the value of wavelengths from the equation (2) in equation (1) we have,
$\text{E = }\dfrac{\text{h }C}{\text{ }\!\!\lambda\!\!\text{ }}$ (3)
Let’s substitute the values in the (3) equation.
\[\text{E = }\dfrac{\text{h }C}{\text{ }\!\!\lambda\!\!\text{ }}\text{ = }\dfrac{(\text{ 6}\text{.626 }\times \text{ 1}{{\text{0}}^{\text{-}}}^{\text{34}}\text{ }\!\!~\!\!\text{ J s})\text{ }(3\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}})}{\text{( }250\text{ }\times \text{ }{{10}^{-9\text{ }}}\text{m) }}\text{ = }\dfrac{1.987\text{ }\times \text{ 1}{{\text{0}}^{-25}}\text{ }}{250\text{ }\times \text{ }{{10}^{-9\text{ }}}}\text{J = 7}\text{.95 }\times \text{ 1}{{\text{0}}^{-19}}\text{ J}\]
Thus, the energy associated with the wavelength, \[\text{ }\!\!\lambda\!\!\text{ }=\text{ }250\text{ }\times \text{ }{{10}^{-9\text{ }}}\text{m }\]is equal to the\[\text{7}\text{.95 }\times \text{ 1}{{\text{0}}^{-19}}\text{ J}\].

Hence, (A) is the correct option.

Note: The planck's constant is related to the momentum of a particle and the wavelength of the radiation. This is called the de Broglie equation. It relates the wave nature of radiation to the particle nature. The relation is stated as follows,
$\text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\text{P}}\text{ = }\dfrac{\text{h}}{\text{mv}}$
Then, the above energy equation can be modified as,
$\text{ E = h C }\times \text{ }\dfrac{\text{m C}}{\text{h}}\text{ = m}{{\text{c}}^{\text{2}}}\text{ }$
This is an Einstein equation of energy.