Question

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1 , (c) an He atom traveling at 1000 m s−1 (a typical speed at room temperature), (d) yourself traveling at 8 km h−1 , and (e) yourself at rest.

Answer 1

Answer:

a)$\lambda=6.63\times10^{-31}m$

b)$\lambda=6.63\times10^{-39}m$

c)$\lambda=9.97\times10^{-11}m$

d)$\lambda=4.03\times10^{-36}$m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ($6.63\times10^{-34}\frac{m^{2}kg}{s}$) and P the momentum of the object that is mass (m) times velocity (v).

a)$\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b)$\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c)$\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d)$\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞