Question

Calculate the minimum amount of energy that the photons must possess to eject electrons from cesium metal. The threshold frequency of cesium metal is 4.6×10^14/s(h=6.63×10^-34Js)

Answer 1

Answer:

The 3.0498 * 10⁻¹⁹ Joules of minimum energy is required to eject electrons from cesium metal.  

Explanation:

Given data:

The threshold frequency of cesium metal, vo= 4.6 * 10¹⁴ s⁻¹

To find: the minimum amount of energy required

We know,  

The minimum energy which is required to eject electrons from Cesium metal is given by

E = h vo

Where, h = Plank's constant = 6.63 * 10⁻³⁴ Js

So, by substituting the value we get

∴ E = [6.63 * 10⁻³⁴ Js] * [4.6 * 10¹⁴ s⁻¹] = 30.498 * 10⁻²⁰ J = 3.0498 * 10⁻¹⁹J

Answer 2

Answer:

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Explanation:

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