Question

when magnetic radiation of wavelength 600 NM falls on the surface of an alkali metal an electron is emitted with a kinetic energy 2.8 ×10^-19 J. the minimum energy needed to remove an electron from the alkali metal is​

Answer 1

Answer:

5×10⁻²⁰J

Explanation:

Note that Ф is the work function of the metal which is the minimum energy required to remove an electron from the metal. Also note that I have used

h = 6.6×10⁻³⁴Js which is an approximate value since you haven't given the value in the sum. Please make sure to substitute the value given if provided in the substitution.

According to the Einstein's Photoelectric Equation,

kmax = hf - Ф

∴ Ф  = hf- kmax  where f=c/λ

now all you have to do is substitute the values to the equation and get the answer

Ф = [6.6×10⁻³⁴Js × (3 × 10⁸ ms⁻¹ / 6×10⁻⁷ m)] -2.8 × 10^-19J

Ф = 3.3×10⁻¹⁹J - 2.8×10⁻¹⁹J

Make sure to ask any part which you don't understand