Double Slit Quantum and Electricity Experiment

[harrietstowe]

Homework Statement

Suppose electrons are sent through a double slit set up with a the slit width = 5.00nm and d the slit separation = 10.0nm. If the maximum current at the screen is 2.30μA, what is the current at the point .500 cm above the central maximum?

Homework Equations

dsinθ=mλ
asinθ=mλ
y=(mλD)/a

The Attempt at a Solution

I am going to assume that the maximum current occurs at the central maximum. I suppose you could find out how many electrons are hitting that region per second by dividing 2.30e-6 by 1.602e-19
Other then that I am stuck.
Thanks

 

[Redbelly98]

For starters, you would want to calculate what angle corresponds to that 0.500 cm distance. However, if they do not tell you how far the screen is from the slits, there is no way to know that.

If you could find that angle, then you could use the formula for the far-field intensity, which is in our library entry on the double slit:

https://www.physicsforums.com/library.php?do=view_item&itemid=203