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A potassium foil is a distance r=3.5m fr...

A potassium foil is a distance `r=3.5m` from an isotropic light source that emits energy at the rate `P=1.5W`. The work function F of potassium is 2.2 eV. Suppose that the energy transported by the incident light were transferred to the target foil continuously and smoothly (that is, if classical physics prevailed instead of quantum physics). How long would it take for the foil to absorb enough energy to eject and electron? Assume that the foil totally absorbs all the energy reaching it can that the to - be-ejected electron collects energy from a circular patch of the foil whose radius is `5.0xx10^(-11)m`, about that of a typical atom.

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To solve the problem, we need to determine how long it takes for the potassium foil to absorb enough energy to eject an electron. We will follow these steps: ### Step 1: Calculate the intensity of the light at the potassium foil The intensity \( I \) at a distance \( r \) from an isotropic point source can be calculated using the formula: \[ I = \frac{P}{A} \] ...
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A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. If power of the source P is 1.5 W and distance of the foil from the source is 3.5 m, the energy received by an electron per second is

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. if work function of the metal of the foil is 2.2 eV, the time taken by electron to come out is nearly