Question

Calculate the energy in kilocalories per mol of the photons of an electromagnetic radiation of wavelength $5700 A^o$.

Answer 1

Hint: Energy of a photon is dependent on the frequency of the light, which is inversely proportional to the wavelength of the light source. One mole is the quantity of particles equivalent to the Avogadro number.

Formulae used: $E = h\nu $
$E = \dfrac{{hc}}{\lambda }$ since $\nu = \dfrac{c}{\lambda }$
$h = 6.62607004 \times {10^{ - 34}}{m^2}kg/s$
$c = 3 \times {10^8}m/s$
${N_A} = 6.022 \times {10^{23}}$
$1cal = 4.184J$

Complete step by step answer:
Using the following equation we can find the energy possessed by a single electron.
$E = h\nu $
Where $E$ is the energy of the photon, $h$ is the Planck’s constant and $\nu $ is the frequency of the emitted light.
$ \Rightarrow $$E = \dfrac{{hc}}{\lambda }$ since $\nu = \dfrac{c}{\lambda }$
Where $c$ is the speed of light in vacuum and $\lambda $ is the wavelength of the light.
The wavelength here is \[5700{A^o}\] which is equal to $5700 \times {10^{ - 10}}m = 5.7 \times {10^{ - 7}}m$ as we know $1{A^o} = {10^{ - 10}}m$. Substituting these values, we get:
$E = \dfrac{{(6.62607004 \times {{10}^{ - 34}}{m^2}kg/s) \times (3 \times {{10}^8}m/s)}}{{5.7 \times {{10}^{ - 7}}m}}$
On solving, we get:
$3.4874 \times {10^{ - 19}}kg{m^2}/{s^2} = 3.4874 \times {10^{ - 19}}J$
Where $J$ is the SI unit of energy, Joule.
We know $1cal = 4.184J$. Therefore $1J = \dfrac{1}{{4.184}}cal$ where $cal$ stands for calorie.
$ \Rightarrow 1J = 0.239cal$
To convert this to kilocalorie ($kcal$), we multiply it with ${10^{ - 3}}$.
$ \Rightarrow 1J = 0.239 \times {10^{ - 3}}kcal$
Therefore, we must now convert the energy value we got in $J$ to $kcal$.
$ \Rightarrow E = 3.4874 \times {10^{ - 19}} \times 0.239 \times {10^{ - 3}}kcal$
On solving this, we get:
$E = 8.335 \times {10^{ - 23}}kcal$
This is the energy possessed by a single photon. To find the energy per mole, we must multiply this value with the Avogadro number.
Avogadro number, ${N_A} = 6.022 \times {10^{23}}$
Therefore, energy per mole,
$E = 8.335 \times {10^{ - 23}} \times 6.022 \times {10^{ - 23}}kcal/mol$
On solving this, we get
$E = 50.193888kcal/mol$

Therefore, the final answer is $E = 50.193888kcal/mol$.

Additional Information: The photon is a type of elementary particle. It is the basic packet of energy, or quantum, of an electromagnetic wave, and acts as the force carrier for the electromagnetic force. Photons do not have any mass. So, in vacuum, they move at the speed of light.

Note: It may be useful to memorize the value of important and widely used constants like the Planck’s constant, as they may not always be mentioned with the question. It is mandatory to remember the value of the Avogadro number as well. In this approach, we have solved the question by converting all the units to SI units. This question can also be solved using the CGS system of units, in which we will directly use calorie instead of first finding the answer in joules. Sometimes, questions of this type can be asked where only the frequency is given. In that case we can directly use the equation $E = h\nu $. Therefore, it is important to clearly understand and memorize the relationships between frequency, wavelength and speed of light.