(a) Use the de Broglie relation λ = h/p to nd the wavelength of a raindrop with mass 1 mg and speed 1 cm/s. Is there a way to set up a raindrop-diraction experiment and detect the wave-like properties of rain?

Question:

a). Use the de Broglie relation λ=h/p to find the wavelength of a raindrop with mass m=1 mg and speed 1cm/s.

ii). Does it seem likely that the wave properties of a raindrop could be easily detected?

b). Find the wavelength of electrons with KE = 500 eV.

c). If a neutron has the same wavelength as blue light (λ=450 nm) what is it's KE?

ii). What if it's an electron?

The answers to the question are

a). The wavelength of the raindrop is 6.626*10⁻²⁶ m

The properties of the rain drop will be hardly detected

b). The wavelength of the electrons is 5.491×10⁻¹¹ m

c). The KE of the neutron is 5.242510⁻²⁸ J

ii). For an electron it will increase to be KE (electron) = 9.6392639×10⁻²⁵ J

Explanation:

Using de Broglie relations, we have

p = h/λ and E = h·f also E = 1/2·m·v²

a). λ= h/p, E= p²/2·m, p = √(2·m·E), λ = h/√(2·m·E)

Where

λ=wavelength

E = energy

p = momentum

m = mass

The kinetic energy of the rain drop  is ×m×v² = 0.5×(‪1×10⁻⁶)(0.01)2

=  5× 10⁻¹¹ J

λ = h/√(2·m·E) = 6.626*10-34 Js/√(2×‪‪1×10⁻⁶×5× 10⁻¹¹)

= 6.626*10⁻²⁶ m

The properties of the rain drop will not be easily detected

b). The electron energy  is equivalent to 500 eV ⇒500 eV × 1.6×10⁻¹⁹ J/eV

= 8×10⁻¹⁷ J

λ = h/√(2·m·E) = 6.626×10⁻³⁴ Js/√(2*×9.1×10⁻³¹×8×10⁻¹⁷)

= 5.491×10⁻¹¹ m

c). λ = h/√(2·m·E) then √(2·m·E) = h/ λ or E = (h/λ)²/(2·m)  

= (6.626×10⁻³⁴/‪5.0×10⁻⁷‬)²/(2×1.674927471×10⁻²⁷)

E = 5.242510⁻²⁸ J

ii). For an electron, we have m = 9.10938356 × 10⁻³¹ kg

λ  = (6.626×10⁻³⁴/‪5.0×10⁻⁷‬)²/(2×9.10938356×10⁻³¹)  = 9.6392639×10⁻²⁵ J

this would be about 43%