Physics, 30.07.2021 20:20 wallsdeandre6927

In a Rutherford scattering experiment, alpha parti- cles having kinetic energy of 7.70 MeV are fired toward a gold nucleus that remains at rest during the collision. The alpha particles come as close as 29.5 fm to the gold nucleus before turning around. (a) Calculate the de Broglie wave- length for the 7.70-MeV alpha particle and compare it with the distance of closest approach, 29.5 fm. (b) Based on this comparison, why is it proper to treat the alpha particle as a particle and not as a wave in the Rutherford scattering experiment

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(a) The de Broglie wavelength is approximately 5.175 × 10⁻¹⁵ meters. The wavelength is lesser than the distance of closest approach

(b) It is proper to treat the alpha particle as a particle and not as wave because the distance of closest approach is much larger than and not comparable to its wavelength for the alpha particle for the alpha particle to be treated as a wave

The given parameters are;

The kinetic energy of the alpha particles = 7.70 MeV = 1.23368 × 10⁻¹² J

The distance from the gold nucleus the alpha particles reach = 29.5 fm

(a) The de Broglie wavelength of a particle is given as follows;

$\mathbf{\lambda = \dfrac{h}{p}}$

Where;

λ = The wavelength

h = Planck's constant = 6.62607004 × 10⁻³⁴ m²·kg/s

p = The momentum of the particle = Mass of an electron, m × Velocity, v

The mass of an alpha particle, m ≈ 6.645 × 10⁻²⁷ kg

Therefore;

$\lambda = \dfrac{h}{m \times v}$

The kinetic energy of the alpha particle, K.E. = (1/2)·m·v²

∴ v = √(2 × K.E./m)

Therefore;

$\lambda = \dfrac{h}{m \times \sqrt{2 \times \dfrac{K.E.}{m} } } = \dfrac{h}{ \sqrt{2 \times m \times K.E.} }$

Plugging in the values of the variables gives;

$\lambda = \dfrac{6.62607004 \times 10 ^{-34} }{ \sqrt{2 \times 6.645 \times 10 ^{-27} \times 1.23368 \times 10^{-12} } } \approx 5.175 \times 10^{-15}$

The de Broglie wavelength of the alpha particle, λ ≈ 5.175 × 10⁻¹⁵ m

The distance of closest approach = 29.5 fm = 29.5 × 10⁻¹⁵ m

Compared to the distance of closest approach, the wavelength of the alpha particle is lesser than the distance of closest approach

(b) Given that the distance of closest approach is six times larger than the wavelength of the alpha particle, and alpha particle behaving as waves are expected to approach closer to the gold nucleus in the region of their wavelength before deflection, therefore, the larger distance of closest approach is indicative of a charged particle to charged particle interaction, and therefore, particle behavior of alpha particles.

Learn more about de Broglie wavelength, particles and wave behavior of radiation here;

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