In a double slit experiment with two slits of width 1.0 um spaced 4um apart, sup

Question

In a double slit experiment with two slits of width 1.0 um spaced 4um apart, suppose a beam of protons is incident on the slits after being accelerated from rest through a potential difference of 2500V.
a) What is the speed of the proton?
b) What is the wavelength of the proton?
c) If the pattern of the detected protons is observed on a fluorescent screen 20m from the slits, what is the center-to-center spacing between the constructive interference maxima?
d) What is the full width of the central diffraction minimum?
e) How many interference fringes will be observed within the central diffraction minimum?

Explanation / Answer

Given       width of slits   W =   1 m                                 = 1 * ( 10 -6 m / m )        distance between two slits   d = 4 m                                                         =   4* ( 10 -6 m / m )        potential difference    V = 2500 V a)    potential energy of proton is                    P.E = V q          here q is charge on proton ( 1.6 *10 -19 C )                   m is mass of proton ( 1.6 *10 -27 kg )                     ( 1/2) m v 2    = ( 2500) ( 1.6 *10 -19 C )                                v 2    =    2 ( 2500) (   ( 1.6 *10 -19 C )   / ( 1.6 *10 -27 )                                 v2    = 5000*108                                    v   = 70.7106 *104   m/s                                         = 7.07106 *105 m/s b)   wave length of proton is                         = h / m v         here   h is planks constant ( 6.635*10-34 Js )                 = ( 6.635*10-34 Js ) /  ( 1.6 *10 -27 ) ( 7.07106 *105 m/s )                        =5.8641*10-11 m      c) distance between screen and slit is   L = 20 cm = 0.2 m         to center spacing between the constructive interference maxima                         y = m L /d                          = (1) ( 5.8641*10-11 m )( 0.2) / 4* ( 10 -6 m )                          = 2.93205 *10-6 m          

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