Question

What minimum speed must an electron have to excite the 492-nm-wavelength blue emission line in the hg spectrum?

Answer 1

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:
$\Delta E = \frac{1}{2} m v^{2}$

Therefore, 
$\frac{1}{2} m v^{2} = \frac{hc}{\lambda}$

And solving for v we get:
$v = \sqrt{ \frac{2hc}{m\lambda} }$

Plugging in numbers (after trasforing into the correct SI units of measurement):
$v = \sqrt{ \frac{(2)(6.6 \cdot 10^{-34})(3 \cdot 10^{8}) }{(9.11 \cdot 10^{-31})(4.92 \cdot 10^{-7}) } }$
=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.