calculate the momentum of a particle ,whose de Broglie's wave length is 0.1nm ?
λ = h/p (deBrogle Equation)
h = Plancks Constant = 6.63 x 10⁻³⁴ Joule-sec
λ = 0.10nm = 1 x 10⁻⁸ meter
p = particle momentum = h/λ = (6.63 x 10⁻³⁴ Joule-sec / 1 x 10⁻⁸ meter) = 6.63 x 10⁻²⁶ Kg·m/sec
λ = h/p (deBrogle Equation)
h = Plancks Constant = 6.63 x 10⁻³⁴ Joule-sec
λ = 0.10nm = 1 x 10⁻10 meter
p = particle momentum = h/λ = (6.63 x 10⁻³⁴ Joule-sec / 1 x 10⁻10 meter) = 6.63 x 10⁻²4 Kg·m/sec
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To calculate the momentum of a particle with a given de Broglie wavelength, we can use the de Broglie equation, which relates the wavelength (λ) to the momentum (p) of a particle:
λ = h/p
Where:
λ = de Broglie wavelength of the particle,
h = Planck's constant (h = 6.626 x 10^-34 Js),
p = momentum of the particle.
In this case, we are given that the de Broglie wavelength (λ) is 0.1 nm.
First, let's convert the wavelength from nanometers (nm) to meters (m). We know that 1 nm is equal to 1 x 10^-9 m.
λ = 0.1 nm = 0.1 x 10^-9 m
Now we can rearrange the de Broglie equation to solve for momentum (p):
p = h/λ
p = (6.626 x 10^-34 Js) / (0.1 x 10^-9 m)
p = 6.626 x 10^-34 / 0.1 x 10^-9
p ≈ 6.626 x 10^-34 / 0.1 x 10^-9
p ≈ 6.626 x 10^-25 kg·m/s
Therefore, the momentum of the particle is approximately 6.626 x 10^-25 kg·m/s.