Question

Calculate energy of one mole of photons of radiation whose frequency is 5 *10^{14} Hz.

NCERT Class XII
Chemistry - Main Course Book I

Chapter 2. Structure of Atom

Answer 1

energy = number of photons * energy in one photon
         = n h v
         = 6.023 * 10²³  *  6.626 * 10⁻³⁴ * 5 * 10¹⁴  Joules
         =  199.54 kJoules

Answer 2

ANSWER:

The energy of one mole of photons of radiation is $\bold{199.51\ \mathrm{KJ} \mathrm{mol}^{-1}}$.

EXPLANATION:

Given:

Frequency $=5 \times 10^{14}\ \mathrm{Hz}$

h = Planck’s constant $=6.626 \times 10^{-34}\ \mathrm{J} / \mathrm{sec}$

Plug the given value of frequency and h constant to calculate the “energy of photon”.

Energy $\bold{=\mathrm{h} \times frequency}$

$\begin{array}{l}{\mathrm{E}=6.626 \times 10^{-34}\ \mathrm{J} / \mathrm{sec} \times 5 \times 10^{14}\ \mathrm{Hz}} \\ {\mathrm{E}=33.13 \times 10^{-20}\ \mathrm{J}=3.313 \times 10^{-19}\ \mathrm{J}}\end{array}$

The “energy of one mole of photon”

$\begin{array}{l}{=\left(3.313 \times 10^{-19} \mathrm{J}\right) \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)} \\ {=199.51\ \mathrm{KJ} \mathrm{mol}^{-1}}\end{array}$