Use the given relationship to calculate the ionization energy of a thallium atom, given that radiation of 58.4 nm produces electrons with a speed of 2310 km⋅s−1. Note that 1 J=1 kg⋅m2⋅s−2.

The ionization energy is 1 × 10^-18 J
The ionization energy is the energy required to remove an electron from an atom.
We have the following information;
wavelength of the photon = 58.4 nm
Speed of the electron = 2310 × 10^3 m/s
Since;
hv = I + 1/2mv^2
v = c/λ
hc/λ = I + 1/2mv^2
I = hc/λ - 1/2mv^2
I = (6.6 × 10^-34 × 3 × 10^8/58.4 × 10^-9) - (1/2 × 9.11 × 10^-31 × (2310 × 10^3)^2)
I = 1 × 10^-18 J
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