Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js is used for the calculation.
Answer:
(a) A virus -------------Classical
(b) A buckyball -----Classical
(c) A mosquito ------ Quantum
(d) A turtle ------------Quantum
Explanation:
Calculating the wavelength using the formula;
λ= h/(mv)
where
λ= Wavelength
h = Planck Constant = 6.62607*10^-9 Js
m = mass in kg
v = velocity in m/s
Virus size = 280. nm = 2.80*10⁻⁷ m
a)
A Virus:
m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg
v = 0.50 µm/s = 5 *10⁻⁷ m/s
h = 6.62607*10^-9 Js
Virus size = 280 nm = 2.80*10⁻⁷ m
Substituting into the formula; we have
λ= h/(mv)
λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)
= 6.62607*10^-9/4.7*10^-26
= 1.4*10^17 m
Classical : Wavelength is bigger than it's size
(b)
A buckyball
m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg
V = 37 m/s
Size = 0.7 nm = 7*10⁻¹⁰ m
Substituting into the formula, we have
λ= h/(mv)
λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)
= 6.62607*10^-9/4.44*10^-23
= 1.49 *10^14 m
Classical : Wavelength is bigger than it's size
(c)
A mosquito
Mass = 1.0 mg = 1*10⁻⁶ kg
v = 1.1 m/s
Size = 6.3 mm = 6.3*10⁻³ m
Substituting into the formula, we have
λ= h/(mv)
λ= 6.62607*10^-9/ ( 1*10⁻⁶* 1.1)
= 6.62607*10^-9/1.1*10^-6
= 6.02*10^-3 m
Quantum Approach: The wavelength and the size are comparable
(d)
A turtle
Mass = 710. g = 0.71 kg
Size = 22. cm = 0.22 m
V = 2.8 cm/s. = 0.028 m/s
Substituting into the formula, we have
λ= h/(mv)
λ= 6.62607*10^-9/ ( 0.71* 0.028)
= 6.62607*10^-9/0.01988
= 3.33*10^-7 m
Quantum Approach: The wavelength and the size are comparable
