Question

9.16 A stream of electrons, each having an energy of 0.5 eV,
impinges on a pair of extremely thin slits separated by 10-2 mm.
What is the distance between adjacent minima on a screen 20 m
behind the slits? (me = 9.108 X 10 kg, 1 eV = 1.602 x 10-19​

Answer 1

distance between adjacent minima is 5m.

it has given that a stream of electrons, each having an energy of 0.5 eV impinges on a pair of extremely thin slits seperated by 10¯² mm.

we have to find the distance between adjacent minima on a screen 20m behind the slits.

first find wavelength of the stream of electrons,

E = hc/λ = 0.5 eV

⇒(6.63 × 10¯³⁴ × 3 × 10^8)/λ = 0.5 × 1.6 × 10^-19 J

⇒λ = (6.63 × 3 × 10^-26)/(0.8 × 10^-19)

= 2.48625 × 10^-6 m

now distance between adjacent minima , ∆y = λD/d

where d is distance between slits and D is distance between slits and screen.

here, D = 20m, d = 10^-2 mm = 10^-5 m

∆y = (2.48625 × 10^-6 × 20)/(10^-5)

= (2.48625 × 2)

= 4.9725 m ≈ 5 m

therefore distance between adjacent minima is 5m