Question

What is the de Broglie wavelength (in nm) of a proton of traveling at 3.18% of the speed of light?

Answer 1

The de Broglie wavelength of a proton traveling at 3.18% of the speed of light is approximately 3.36 × 10^-9 nm.

Step-by-step explanation:

The de Broglie wavelength of a particle is given by the equation:

λ = h / (mv)

Where:

• λ is the de Broglie wavelength
• h is Planck's constant (6.626 × 10^-34 Js)
• m is the mass of the particle (in this case, the mass of a proton is approximately 1.67 × 10^-27 kg)
• v is the velocity of the particle

To find the de Broglie wavelength of a proton traveling at 3.18% of the speed of light, we can calculate the velocity of the proton using the formula:

v = βc

Where:

• β is the fraction of the speed of light (in this case, 3.18% or 0.0318)
• c is the speed of light (3.00 × 10^8 m/s)

Substituting the values into the equations, we have:

v = 0.0318 × 3.00 × 10^8 m/s

v ≈ 9.54 × 10^6 m/s

Now we can calculate the de Broglie wavelength:

λ = (6.626 × 10^-34 Js) / (1.67 × 10^-27 kg × 9.54 × 10^6 m/s)

λ ≈ 3.36 × 10^-12 m

Converting the wavelength to nanometers:

λ ≈ 3.36 × 10^-9 nm

Answer 2

Answer is: de Broglie wavelength of a proton is 3,4·10⁻⁵ nm.
v(proton) = 0,038 · 3·10⁸ m/s.
v(proton) = 1,14·10⁷ m/s; speed of proton.
m(proton) = 1,67·10⁻²⁷ kg.
h = 6,62607004·10⁻³⁴ m²·kg/s; Planck constant.
λ(proton) = h / m(proton) · v(proton).
λ(proton) = 6,62607004·10⁻³⁴ m²·kg/s ÷ (1,67·10⁻²⁷ kg · 1,14·10⁷ m/s).
λ(proton) = 3,48·10⁻¹⁴ m · 10⁹ nm/m = 3,4·10⁻⁵ nm.