Question

Calculate the energy in kilocalorie per mole of the photons of an electromagnetic radiation of wavelength 7600 angstrom. Answer should be in kcal per mole.

Answer 1

Here's the solution. Hope it helps.

Answer 2

Given that,

The wavelength of an electromagnetic radiation is 7600 A.

To find,

The energy in kilocalorie per mole of the photons.

Solution,

The energy for one photon is given by :

$E=\dfrac{hc}{\lambda}$

For one mole of photons,

$E=\dfrac{N_Ahc}{\lambda}$, $N_A$ = Avogadro's constant

So,

$E=\dfrac{6.022\times 10^{23}\times 6.62\times 10^{-34}\times 3\times 10^8}{7600\times 10^{-10}}\\\\E=1.57\times 10^5\ J/mol$

We know that,

1 calorie = 4.2 J

So,

$E=\dfrac{1.575\times 10^5}{4.2}\\\\E=37500\ cal/mol\\\\E=37.5\ kcal/mol$

So, the energy of the photons is 37.5 kcal/mol.