Question

How can the ionization energy of sodium be calculated knowing that the value of the wavelength at the Start of the continuum in the emotion spectrum is 242nm?

Answer 1

Hint: The value of the wavelength at the start of the continuum represents the ionization energy: the energy needed to remove a 3s electron from a sodium atom (the red arrow in the diagram below).

Complete step by step solution:
Wavelength $ \lambda = 242nm = 2.42 \times {10^{ - 7}}m $ We know,
Energy $ (E) = hc/\lambda \; $
Here, h is Planck's constant
C is the speed of light
And λ is the wavelength of photons
 $ so,E = \dfrac{{6.625 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m/s}}{{2.421 \times {{10}^{ - 7}}m}} $
 $ = 0.0821 \times {10^{ - 17}}\;J{\text{ }}/{\text{ }}atom\; $
 $ 0.0821 \times {10^{ - 17}} $ J energy is sufficient for ionization of one Na atom, so it is the ionization energy of Na .
 $
  hence,\;I.E = 0.0821 \times {10^{ - 17}}\;J/atom \\
  \; = 0.0821 \times {10^{ - 17}} \times 6.022 \times {10^{22}}\;J/mol\; \\
   = 493.5\;KJ/mol\; \\
  $
Assume we have a compartment of hydrogen gas through which an entire arrangement of photons is passing, permitting numerous electrons to climb to more significant levels. At the point when we turn off the light source, these electrons "fall" back down from bigger to more modest circles and radiate photons of light—at the same time, once more, just light of those energies or frequencies that relate to the energy contrast between passable circles. The orbital changes of hydrogen electrons that offer ascent to some otherworldly lines have appeared.

Note:
How about we take a gander at the hydrogen molecule from the point of view of the Bohr model. Assume a light emission light (which comprises photons of every obvious frequency) radiates through a gas of nuclear hydrogen.