A) Suppose you perform a double slit experiment with electrons of KE = 25 eV. Th

Question

A) Suppose you perform a double slit experiment with electrons of KE = 25 eV. The screen is located L=10 m behind the double slit. You aim to separate the central maximum and the first maximum to the side by x=1 mm. What must be the spacing d between the two slits?

B) Suppose you perform a double slit experiment with C60 molecules of KE = 25 eV. The screen is located L=10 m behind the double slit. You aim to separate the central maximum and the first maximum to the side by x=1 mm. What must be the spacing d between the two slits?

C) Suppose you perform a double slit experiment with 5g-bullets at a velocity of 120 m/s. The screen is located L=10 m behind the double slit. You aim to separate the central maximum and the first maximum to the side by x=1 mm. What must be the spacing d between the two slits?

Explanation / Answer

here,

PART A:

KE = E = hc/L

wavelength L = hc/E

L = 6.626*10^-34 * 3*10^8/(25*1.6 *10^-19)

L = 49.65 nm

Y = mLR/d

here m = order of the fringe = 1

R = 10 m

d = ?

Y = 1 mm

so

d = (1 * 49.65 *10^-9 * 10)/(1*10^-3)

d = 0.49 mm (answer)

------------------------------

part B :

KE = E = hc/L

wavelength L = hc/E

L = 6.626*10^-34 * 3*10^8/(25*1.6 *10^-19)

L = 49.65 nm

Y = mLR/d

here m = order of the fringe = 1

R = 10 m

d = ?

Y = 1 mm

so

d = (1 * 49.65 *10^-9 * 10)/(1*10^-3)

d = 0.49 mm (answer)

---------------------------------------

part C:

KE = 0.5 mv^2

KE = 0.5* 5 e -3 * 120*120

KE = 36 J

waveelngth   = hc/E

L = (6.626*10^-34 * 3*10^8)/(36)

L = 5.52 *10^-27 m

so

d = (5.52*10^-27 * 1* 10/(1*10^-3)

d = 5.52*10^-23 m (answer)

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