Question

Consider a double slit apparatus that produces several dark minima. The third from the center dark minimum is at 45.0° for 635-nm light.

a.)What is the separation between the double slits, in meters?

b.)What slit separation, in meters, would be needed to produce the same pattern for 1.00-keV protons?

c.)What is the problem with extending the results from part (a) to part (b)?

a. Protons with that much energy will not diffract.

b. Light is a wave, but a proton is a particle.

c. Protons cannot have that much energy.

d. Protons do not diffract like light.

e. The slit in part (b) would have to be smaller than the usual distance between atoms.

Answer 1

given

theta = 45 degrees
lamda = 635 nm

a) let d is the seoaration between the doule slits.

for mth dark fringe, d*sin(theta) = (m - 1/2)*lamda

d*sin(45) = (3 - 1/2)*lamda

d = 2.5*635*10^-9/sin(45)

= 2.24*10^-6 m <<<<<<<<-----------------Answer

b) kinetic energy of protons, KE = 1 keV

wavelength of protons, lamda = h/p

= h/sqrt(2*m*KE)

= 6.626*10^-34/sqrt(2*1.67*10^-27*1*10^3*1.6*10^-19)

= 9.06*10^-13 m

so, d = 2.5*lamda/sin(45)

= 2.5*9.06*10^-13/sin(45)

= 3.20*10^-12 m <<<<<<<<-----------------Answer

c) e. The slit in part (b) would have to be smaller than the usual distance between atoms.

because, 3.2*10^-12 m << 10^-10 m