(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

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

i think the answer is a, hope this is right, that would be my best guess.