Respuesta :
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?
Answer:
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 [tex]\frac{1}{2}[/tex]×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
