Question

When light with a wavelength of 58.4nm was applied to a rubidium atom, electrons with a velocity of 2450 `kms^-1` were emitted. What is the ionization energy of a rubidium atom? (1J= `1kgm^2s^-2`)

Answer 1

I tried this but check it anyway...

Explanation:

Ok, I think that the energy of your photon of light (`E=hf` Panck's Constant times frequency) will be used to:

1) free the electron form the atom=ionization energy=`x`;
2) give the electron Kinetic Energy: `K=1/2mv^2`.

[I considered the high speed to be Relativisticly affected but after considering the Lorents Term `gamma-1/sqrt(1-v^2/c^2)~~1` I think it isn't]

so we get:

`E=K+x`

`hf=1/2mv^2+x`

rearrange:

`x=hf-1/2mv^2`

but: `c=lambdaf`

and so: `f=c/lambda`

giving:

`x=hc/lambda-1/2mv^2`

using the mass for the electron, Planck's Constant, the speed of light from literature and our data:

`x=(6.63xx10^-34*(3xx10^8)/(58.4xx10^-9))-[1/2*9.1xx10^-31*(2.45xx10^6)^2]=6.75xx10^-19J`

It kind of fits with the `4.17eV` given in:

https://webbook.nist.gov/cgi/cbook.cgi?ID=C7440177&Mask=864