Question

The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 9.00×1010 atoms of zinc emitting light in the instrument flame per second, what energy (in joules) must the flame supply during this time to achieve this level of emission?

Answer 1

Answer:

$8.4\cdot 10^{-8}J$

Explanation:

The energy emitted by a single photon is given by:

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

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

For the photons emitted by the zinc atoms,

$\lambda = 214 nm = 214 \cdot 10^{-9} m$

So the energy of a single photon emitted is

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{214\cdot 10^{-9}}=9.3\cdot 10^{-19}J$

And since the number of atoms is

$N=9.0\cdot 10^{10}$

The total energy emitted will be

$E=NE_1 = (9.0\cdot 10^{10})(9.3\cdot 10^{-19})=8.4\cdot 10^{-8}J$