Double Slit Quantum and Electricity Experiment

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SUMMARY

The discussion centers on calculating the current at a specific point above the central maximum in a double slit experiment involving electrons. Given a slit width of 5.00 nm and a slit separation of 10.0 nm, the maximum current at the screen is 2.30 μA. To find the current at a point 0.500 cm above the central maximum, participants suggest calculating the corresponding angle and using the far-field intensity formula. The lack of distance information from the slits to the screen presents a challenge in determining the angle accurately.

PREREQUISITES
  • Understanding of double slit experiment principles
  • Familiarity with wave-particle duality of electrons
  • Knowledge of the equations dsinθ=mλ and y=(mλD)/a
  • Basic proficiency in calculating electric current and charge of an electron
NEXT STEPS
  • Calculate the angle corresponding to a 0.500 cm distance using the double slit equations
  • Research the far-field intensity formula for double slit experiments
  • Explore the impact of slit width and separation on interference patterns
  • Study the relationship between current and electron flow in quantum experiments
USEFUL FOR

Students and educators in physics, particularly those focusing on quantum mechanics and experimental physics, as well as anyone interested in the principles of electron behavior in double slit experiments.

harrietstowe
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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
 
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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
 

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