**How can the ionization energy**

**How can the ionization energy**

#### How can the ionization energy of rubidium per atom be calculated if light of wavelength 5.84 x 10-8m produces electrons with a speed of 2.450 x 10 6 ms-1?

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Photon energy = Ionization Energy + Kinetic Energy

We get the energy of photon from the Plank’s equation, E = h f, where f is the frequency of the photon and h is the Plank Constant which is equal to 6.626 * 10^-34 Js

The kinetic energy of an electron is KE = 1/2 m v^2, where m is the mass, i. e., 9.11 * 10^-31 kg and the velocity v is 2.45 * 10^6 m/s

So it can be said that,

h f = IE + 1/2 m v^2

or, IE = h f – 1/2 m v^2

As c = f * λ, we can say that, IE = (h c) / λ – 1/2 m v^2

Putting the values we get, that,

IE = (6.626 * 10^-34 * 3 * 10^8) / (5.84 *10^-8) – {1/2 * 9.11 * 10^-31 * (2.45 * 10^6)^2}

= (3.404 * 10^-18) – (2.734 * 10^-18) J

= 6.7 * 10^-19 J

This is the ionization energy of an atom of Rubidium.