When light of 470 nm falls on the surface of potassium metal, electrons are emitted with a velocity of `6.4 xx 10^(4)ms^(-1)`. What is the minimum energy required to remove one moles electrons from potassium metal?

Explanation

To solve the problem of finding the minimum energy required to remove one mole of electrons from potassium metal when light of 470 nm falls on its surface, we can follow these steps:

### Step 1: Calculate the Kinetic Energy of the Emitted Electrons
The kinetic energy (KE) of the emitted electrons can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
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