Question

The energy difference between the 5d and the 6s sublevels in gold accounts for its color. If this energy difference is about 2.7 eV (electron volt; 1 eV = 1.602 × 10−19 J), calculate the wavelength of light absorbed in the transition of an electron from the 5d subshell to the 6s subshell. Round the answer to the correct number of significant figures.

Answer 1

Answer:

The wavelength of light absorbed in the transition is 459 nm.

Explanation:

Energy difference between 5-d and the 6-s sub-levels in gold = ΔE

$\Delta E=2.7 eV=2.7 eV\times 1.602\times 10^{-19} J=4.3254\times 10^{-19} J$

Let the wavelength of light absorbed in the transition 5-d to 6-s be $\lambda$

The relation between energy and wavelength is given by:

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon of the light

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of the photon

$4.3254\times 10^{-19} J=\frac{6.63\times 10^{-34}Js\times 3\times 10^8 m/s}{\Lambda }$

$\Lambda =4.59\times 10^{-7} m = 459 nm$

$1nm = 10^{-9 } nm$

The wavelength of light absorbed in the transition is 459 nm.