A beam of electrons, each with the same kinetic energy, illuminates a pair of sl

Question

A beam of electrons, each with the same kinetic energy, illuminates a pair of slits separated by a distance of 58 nm. The beam forms bright and dark fringes on a screen located a distance 2.6 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference experiment. The bright fringes are found to be separated by a distance of 0.6 mm. What is the kinetic energy of the electrons in the beam? Planck's constant is 6.63 times 10^34 J middot s. Answer in units of keV.

Explanation / Answer

sepration between slits , d = 58 nm

d = 5.8 * 10^-8 m

screen distance , D = 2.6 m

y = 0.6 * 10^-3 m

let the wave length be lamda

y = lamda * D/d

0.6 * 10^-3 = lamda * 2.6 /( 5.8 * 10^-8)

lamda = 1.34 * 10^-11 m

the kinetic energy of the electrons , KE = h * c/lamda

KE = 6.626 * 10^-34 * 3 * 10^8 /( 1.34 * 10^-11)

KE = 1.485 * 10^-15 J

KE = 1.485/1.6 * 10^4 eV

KE = 9.28 * 10^3 eV

KE = 9.28 KeV

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