# Quantum Chemistry

Chapter01 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29

Exercise. This is the solution to exercise 1.1 in the book.

Solution. Recall that:

$E=h\nu =\frac{hc}{\lambda }$

where $h$ is Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the photons with energy $E$. Hence:

$E=\frac{6.626\cdot {10}^{-34}J\cdot s\cdot 3\cdot {10}^{8}m∕s}{1064\cdot {10}^{-9}m}=1.868\cdot {10}^{-19}J=1.166eV$

The energy of the pulse is:

${E}_{p}=P\cdot \tau =5\cdot {10}^{6}W\cdot 2\cdot {10}^{-8}s=0.1J$

Hence the number of photons of wavelength 1064-nm contained in the pulse is:

$N=\frac{{E}_{p}}{E}=\frac{0.1J}{1.868\cdot {10}^{-19}J}=5.35\cdot {10}^{17}$