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

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

Step-by-step 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 =
`(1)/(2) mv^(2)`

Where
`m` is mass

and
`v` is velocity

Hence, we can write that

`hf = I.E + (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 = (c)/(\lambda)`

Then,

`(hc)/(\lambda) = I.E + (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,

`(hc)/(\lambda) = I.E + (1)/(2)mv^(2)`

Then,

`(6.626* 10^(-34) * 3.0 * 10^(8) )/(225 * 10^(-9) ) = 8.239 * 10^(-19) + (1)/(2)(9.109*10^(-31) )v^(2)`

`8.835 * 10^(-19) = 8.239 * 10^(-19) + 4.5545 * 10^(-31)v^(2)`
`8.835 * 10^(-19) - 8.239 * 10^(-19) = 4.5545 * 10^(-31)v^(2)`

`5.96 * 10^(-20) = 4.5545 * 10^(-31)v^(2)`

`v^(2) = (5.96 * 10^(-20))/(4.5545 * 10^(-31))`

`v^(2) = 1.3086 * 10^(11)`

`v = \sqrt{1.3086 * 10^(11) }`

`v = 361745.77 m/s`

`v = 3.6 * 10^(5) m/s`

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