Question

A laser with a wavelength of 225 nm is shown on an isolated gas-phase sodium atom. Calculate the velocity of the ejected electron from the ionized atom. The first ionisation energy of sodium is 496 kJ/mol and the mass of an electron is 9.109 x 10^-31 kg.

Answer 1

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 = $hf$

Where $h$ is Planck's constant ( $h$ = 6.626 × 10⁻³⁴ kgm²/s)

and $f$ is the frequency

Also,

Kinetic energy, K.E = $\frac{1}{2} mv^{2}$

Where $m$ is mass

and $v$ is velocity

Hence, we can write that

$hf = I.E + \frac{1}{2}mv^{2}$

But, $c = f\lambda$

where $c$ is the speed of light ( $c$ = 3.0 × 10⁸ m/s)

and λ is the wavelength

∴ $f = \frac{c}{\lambda}$

Then,

$\frac{hc}{\lambda} = I.E + \frac{1}{2}mv^{2}$

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,

$\frac{hc}{\lambda} = I.E + \frac{1}{2}mv^{2}$

Then,

$\frac{6.626\times 10^{-34} \times 3.0 \times 10^{8} }{225 \times 10^{-9} } = 8.239 \times 10^{-19} + \frac{1}{2}(9.109\times10^{-31} )v^{2}$

$8.835 \times 10^{-19} = 8.239 \times 10^{-19} + 4.5545 \times 10^{-31}v^{2}$$8.835 \times 10^{-19} - 8.239 \times 10^{-19} = 4.5545 \times 10^{-31}v^{2}$

$5.96 \times 10^{-20} = 4.5545 \times 10^{-31}v^{2}$

$v^{2} = \frac{5.96 \times 10^{-20}}{4.5545 \times 10^{-31}}$

$v^{2} = 1.3086 \times 10^{11}$

$v = \sqrt{1.3086 \times 10^{11} }$

$v = 361745.77 m/s$

$v = 3.6 \times 10^{5} m/s$

Hence, the velocity of the ejected electron from the ionized atom is 3.6 × 10⁵ m/s