Answer:
The velocity of the ejected electron from the ionized atom is 3.6 × 10⁵ m/s
Explanation:
Using the conservation of energy, we can write that
Photon energy (E) = Ionisation energy (I.E) + Kinetic energy (K.E)
Photon energy, E = [tex]hf[/tex]
Where [tex]h[/tex] is Planck's constant ( [tex]h[/tex] = 6.626 × 10⁻³⁴ kgm²/s)
and [tex]f[/tex] is the frequency
Also,
Kinetic energy, K.E = [tex]\frac{1}{2} mv^{2}[/tex]
Where [tex]m[/tex] is mass
and [tex]v[/tex] is velocity
Hence, we can write that
[tex]hf = I.E + \frac{1}{2}mv^{2}[/tex]
But, [tex]c = f\lambda[/tex]
where [tex]c[/tex] is the speed of light ( [tex]c[/tex] = 3.0 × 10⁸ m/s)
and λ is the wavelength
∴ [tex]f = \frac{c}{\lambda}[/tex]
Then,
[tex]\frac{hc}{\lambda} = I.E + \frac{1}{2}mv^{2}[/tex]
From the question, the first ionisation energy of sodium is 496 kJ/mol
This is the ionisation energy for 1 mole of sodium,
For 1 atom of sodium, we will divide by Avogadro's constant
∴ The ionisation energy becomes
(496 KJ/mol) / (6.02 × 10²³ molecules)
= 8.239 × 10⁻¹⁹ J
This is the ionisation energy for one atom of sodium
Now, to determine the velocity of the ejected electron from the ionized atom,
From,
[tex]\frac{hc}{\lambda} = I.E + \frac{1}{2}mv^{2}[/tex]
Then,
[tex]\frac{6.626\times 10^{-34} \times 3.0 \times 10^{8} }{225 \times 10^{-9} } = 8.239 \times 10^{-19} + \frac{1}{2}(9.109\times10^{-31} )v^{2}[/tex]
[tex]8.835 \times 10^{-19} = 8.239 \times 10^{-19} + 4.5545 \times 10^{-31}v^{2}[/tex][tex]8.835 \times 10^{-19} - 8.239 \times 10^{-19} = 4.5545 \times 10^{-31}v^{2}[/tex]
[tex]5.96 \times 10^{-20} = 4.5545 \times 10^{-31}v^{2}[/tex]
[tex]v^{2} = \frac{5.96 \times 10^{-20}}{4.5545 \times 10^{-31}}[/tex]
[tex]v^{2} = 1.3086 \times 10^{11}[/tex]
[tex]v = \sqrt{1.3086 \times 10^{11} }[/tex]
[tex]v = 361745.77 m/s[/tex]
[tex]v = 3.6 \times 10^{5} m/s[/tex]
Hence, the velocity of the ejected electron from the ionized atom is 3.6 × 10⁵ m/s
