Question

A beam of C60 molecules (basically a molecule-sized ball that is 7.0 × 10^−10 m in diameter and has a mass of 1.2 × 10^−24 kg ) is directed at two small slits in a very thin metal sheet. The
molecules in the beam are travelling at 1 cm/s and the apparatus is in a vacuum. We would like to use this beam + slits to observe the wave properties of the molecules via diffraction.

a) What is the de Broglie wavelength of the molecules in the beam? (ANS: 5.5 x 10^-8 m)

b) Explain / calculate what size and spacing should be used for the slits in order to observe a
clear interference pattern on a screen 1 m away. (Hint: Depends on your assumptions.... but two slits, each 10 x 10^-6 m wide, spaced 55 x 10^-6 m apart would result in interference maxima 1 mm apart on the screen (which should be easily measurable).)

Answer 1

Answer on Question #43164, Physics, Nuclear Physics

Task:

A beam of C60 molecules (basically a molecule-sized ball that is $7.0 \times 10^{\wedge} - 10\mathrm{m}$ in diameter and has a mass of $1.2 \times 10^{\wedge} - 24\mathrm{kg}$ ) is directed at two small slits in a very thin metal sheet. The molecules in the beam are travelling at $1\mathrm{cm/s}$ and the apparatus is in a vacuum. We would like to use this beam + slits to observe the wave properties of the molecules via diffraction.

a) What is the de Broglie wavelength of the molecules in the beam? (ANS: $5.5 \times 10^{\wedge} - 8 \, \text{m}$ )

b) Explain / calculate what size and spacing should be used for the slits in order to observe a clear interference pattern on a screen 1 m away. (Hint: Depends on your assumptions... but two slits, each $10 \times 10^{\wedge} - 6$ m wide, spaced $55 \times 10^{\wedge} - 6$ m apart would result in interference maxima 1 mm apart on the screen (which should be easily measurable).)

Solution:

a) the de Broglie wavelength of the molecules in the beam :

$$\lambda_ {B} = \frac {h}{p} = \frac {h}{m V} = \frac {6 . 6 2 \cdot 1 0 ^ {- 3 4}}{1 . 2 \cdot 1 0 ^ {- 2 4} \cdot 1 0 ^ {- 2}} \approx 5. 5 \cdot 1 0 ^ {- 8} m.$$

b)

$$d \sin \varphi = m \lambda ; m = 1$$ $$\sin \varphi \approx t g \varphi = a / L;$$ $$\frac {\lambda L}{d} < 1 0 ^ {- 3} \quad \text {a - d i s t a n c e b e t w e e n i n t e r f e r e n c e m a x i m u m s , L = 1 m .}$$ $$\frac {\lambda L}{b + c} < 1 0 ^ {- 3} \Rightarrow b > 5. 5 \cdot 1 0 ^ {- 5} - c$$

d- period diffraction grating

c-slits width

b-distance between the slits

So if $c = 10 \cdot 10^{-6} \, \text{m}$ , then $b > 44 \cdot 10^{-6} \, \text{m}$ for example $b = 55 \cdot 10^{-6} \, \text{m}$ .

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