Answer to Question #145543 in General Chemistry for gurleen

Question

A photon of light with a frequency of 6.9254x 10 14 s -1 energizes an electron so it jumps
from one orbit to another. What is the wavelength of the energizing wave? What colour
of the electromagnetic spectrum does it have? Which series of spectral waves does it
belong to? Calculate the energy levels it jumps from and to. [6 marks]

Answer 1

Frequency$(f)=6.9254\times 10^{14}s^{-1}$

Speed of light $c=3.0\times 10^8m/s$

Conversation factor$1m=10^9nm$

Wavelength$\lambda=unknown$

From the formula;

$c=$ $\lambda c$

$\lambda=$ $c\over f$

$\lambda=$ $($$3.0\times 10^8ms^{-1}\over6.9254\times10^{14}s^{-1}$ $)$ $\times ($ $1m\over 10^9nm$ $)$

$=4.33\times 10^{-17}nm$

The frequency $(f)=6.9254\times 10^{14}s^{-1}$ and $\lambda=$ $4.33\times 10^{-17}nm$

From the Electromagnetic spectrum, with respect to wavelength, the color$= violet$

Series of spectral waves$= gamma rays$

The energy level$(E)$ $=unknown$

From the formula;

$E=hf$

Where $E= Energy(eV)$

$h=$ Planks constant$(4.14\times 10^{-15}eVs^{-1}$ $)$

$c=$ speed of light$(3.0\times 10^8ms^{-1}$

$\therefore$ $E=(4.14\times 10^{-15}eVs^{-1})$ $\times$ $(3.0\times 10^8ms^{-1})$

$=2.867eV$