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.

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 =

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

and is the frequency

Also,

Kinetic energy, K.E =

Where is mass

and is velocity

Hence, we can write that

But,

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

and λ is the wavelength

∴

Then,

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,

Then,

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

the answer is a and have a great day