A beam of electrons with energy 48 kV is shot through two narrow slitsin a barri

Question

A beam of electrons with energy 48 kV is shot through two narrow slitsin a barrier. The slits are a distance 1.8E-6 m apart. If a screen is placed33.0 cm behind the barrier,calculate the spacing between the “bright” fringes ofthe interference pattern produced on the screen. I am not getting the right answer with doing it the way theysaid to. Thanks! A beam of electrons with energy 48 kV is shot through two narrow slitsin a barrier. The slits are a distance 1.8E-6 m apart. If a screen is placed33.0 cm behind the barrier,calculate the spacing between the “bright” fringes ofthe interference pattern produced on the screen. I am not getting the right answer with doing it the way theysaid to. Thanks!

Explanation / Answer

Have you tried n = xd/L Where:      is the wavelength      n is the order of maximum observed(central maximum is n=0)     d is the separation of the slits, thedistance between A and B in the diagram to the right     x is the distance between the bands of lightand the central maximum (also called fringe distance),     L is the distance from the slits to thescreen centerpoint. To find the wavelength = c/f    where:                f- frequency               c - the speed of light In terms of energy E and Plank's constant h we have f= h/E now = cE/h and    n = xd/L becomes n cE/h= xd/L x   =n c E L/(h d) x= x2- x1 so for n=1 and n=2 we have x= 2 c E L/(h d) - c E L/(h d) x= c E L/(h d) Just plug and paly Please let me know if you have any questions.

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