Calculate the kinetic energy of the ejected electron when ultra-violet radiation of frequency `1.6 xx 10^(15) s^(-1)` strikes the surface of potassium metal. Threshold frequency of potassium is `5 xx 10^(14) s^(-1) (h = 6.63 xx 10^(-34) Js)`.

Explanation

To calculate the kinetic energy of the ejected electron when ultraviolet radiation strikes the surface of potassium metal, we can use the photoelectric effect equation:

\[ KE_{max} = h \nu - h \nu_0 \]

Where:
- \( KE_{max} \) is the maximum kinetic energy of the ejected electron.
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)).
- \( \nu \) is the frequency of the incident radiation (\( 1.6 \times 10^{15} \, \text{s}^{-1} \)).
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