The energy required by the excitation of the line is: ΔE = hν = hc / λ where: ΔE = energy difference h = Planck constant ν = line frequency c = speed of light λ = line wavelength
The energy difference must be supplied by the electron, supposing it transfers all its kinetic energy to excite the line: [tex]\Delta E = \frac{1}{2} m v^{2} [/tex]
Therefore, [tex]\frac{1}{2} m v^{2} = \frac{hc}{\lambda} [/tex]
And solving for v we get: [tex]v = \sqrt{ \frac{2hc}{m\lambda} } [/tex]
Plugging in numbers (after trasforing into the correct SI units of measurement): [tex]v = \sqrt{ \frac{(2)(6.6 \cdot 10^{-34})(3 \cdot 10^{8}) }{(9.11 \cdot 10^{-31})(4.92 \cdot 10^{-7}) } }[/tex] =9.4 · 10⁵ m/s
Hence, the electron must have a speed of 9.4 · 10⁵ m/s in order to excite the 492nm line.
