A beam of electrons with a speed of is incident on a slit of width 200. nm. The distance to the detector plane is chosen such that the distance between the central maximum of the diffraction pattern and the first diffraction minimum is 0.300 cm. How far is the detector plane from the slit?

  L =( 5.17 10⁻⁶ v) m

Explanation:

This is a diffraction exercise, which is described by the equation

           a sin θ = n λ

where a is the width of the slit a = 200 nm = 200 10⁻⁹ m, n is the order of diffraction n = 1 and lan the wavelength in this case we use the DeBroglie relation to find the wavelength

                 λ  = h´ / p = h´ / mv

with h´ = h / 2π

let's use trigonometry for distance,

            sin θ = y / L

             tan θ = sin θ / cos θ

since these experiments the angle is very small, we approximate

            tan θ = sin θ = y / L

we substitute

             a y / L = n  λ  

             L = a y / n  λ  

             L = a y 2π mv / n h

we calculate

             L = 200 10⁻⁹ 0.300 10⁻² 2π 9.1 10⁻³¹ v / (1   6.63 10⁻³⁴)

             L = 517.44 10⁻⁸ v m

             L =( 5.17 10⁻⁶ v) m

for a specific value we must know the speed of the electrons

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards