Question

Question #58a54

Answer 1

`sf(6.10xx10^(17)` photons strike the leaf every second.

Explanation:

Firstly we need to find the number of Joules of energy incident on the leaf in 1 second.

Then we find the energy of 1 photon.

Then we can find how many of these photons we will need to make up the required energy.

To get consistent units we need to find the area of the leaf in `sf(m^2)`.

`sf(1color(white)(x)cm^2=10^(-4)color(white)(x)m^2)`

So the area of the leaf is given by:

`sf(A=2.25xx10^(-4)color(white)(x)m^2)`

A Watt is 1 Joule per second so the amount of energy incident on the leaf in 1 second is given by:

`sf(E=1070xx2.25xx10^(-4)=0.24075color(white)(x)J)`

To find the energy of 1 photon we use the Planck expression:

`sf(E_("photon")=hf=(hc)/lambda)`

`sf(E_("photon")=(6.626xx10^(-34)xx3.00xx10^(-8))/(504xx10^(-9))=0.03944xx10^(-17)color(white)(x)J)`

So the number of photons required to deliver 0.24075 J of energy in 1 second is given by:

`sf(n=(0.24075)/(0.03944xx10^(-17))=6.10xx10^(17))`