Answer :
[tex]7.06 \times 10^{-36} \mathrm{m}[/tex] is the de broglie wavelength for the given data.
Explanation:
According to the duality of wave particles, this de Broglie wavelengths present the wavelengths in all quantum mechanics objects and evaluates the probability density for determining an object at specific points in the space configuration. It also states that particles can have wave properties.
Its equation used to define the matter’ wave properties, especially the electron wave properties. It can be expressed by the following equation,
[tex]\lambda=\frac{h}{m \times v}[/tex]
Where,
[tex]\lambda[/tex] = - The wavelength
h - The Planck constant ( [tex]6.626176 \times 10^{-34} \text { joule }-\text { seconds }[/tex])
m - The mass
v – velocity
Given data:
Mass (m) = 67 kg
v = 1.4 m/s
To find: de Broglie wavelength (λ)
By substituting the given values in the above equation, we get
[tex]\lambda=\frac{6.626176 \times 10^{-34}}{67 \times 1.4}=\frac{6.626176 \times 10^{-34}}{93.8}=0.0706 \times 10^{-34}[/tex]
When multiply and divide by 100, we get
[tex]\lambda=7.06 \times 10^{-36} \mathrm{m}[/tex]
