Question

Q10) Calculate the momentum of a particle which has a de Broglie wavelength of 0.1 nm.

Answer 1

we know that,lambs=h/mv

mv=h/lambda

mv=6.62×10^-34/0.1×10^-9

mv=66.2×10^-25

hope it helps.

Answer 2

Hi there.

Given Conditions ⇒ 

Broglie Wavelength (λ) = 0.1 nm.
∵ 1 nm = 10⁻⁹ m.
∴ 0.1 nm = 10⁻¹⁰ m.

Also, Plant's Constant = 6.62 × 10⁻³⁴ J-s.

Now, Using the Formula,

Wavelength = Plant's Constant/Momentum.
∴ Momentum = Plank's constant/Wavelength.
∴ Momentum = 6.62 × 10⁻³⁴/10⁻¹⁰
⇒ Momentum = 6.62 × 10⁻²⁴ kg m/s.


Hence, the momentum of the Particles is 6.62 × 10⁻²⁴ kg m/s.


Hope it helps.