Question

Assuming that the smallest measurable wavelength in an experiment is 0.950 fm , what is the maximum mass of an object traveling at 241 m ⋅ s − 1 for which the de Broglie wavelength is observable?

Answer 1

Answer:

The maximum mass is $2.89x10^{-21}Kg$

Explanation:

The wavelength of the electron can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$  

$m = \frac{h}{\lambda v}$  (2)

Where m is the mass and v is the velocity.

Before using equation 2 it is necessary to express the wavelength from femtometers to meters.

$\lambda = 0.950fm .\frac{1m}{1x10^{15}fm}$ -- $9.5x10^{-16}m$

Finally, equation 2 can be used.

$m = \frac{6.624x10^{-34} J.s}{(9.5x10^{-16}m)(241m/s)}$

But $1J = Kg.m^{2}/s^{2}$

$m = \frac{6.624x10^{-34} Kg.m^{2}/s^{2}.s}{(9.5x10^{-16}m)(241m/s)}$

$m = 2.89x10^{-21}Kg$    

Hence, the maximum mass is $2.89x10^{-21}Kg$

Answer 2

Answer:

0.0289 x 10⁻¹⁹kg

Explanation:

The de Broglie wavelength, λ, of a particle varies inversely with the momentum (the product of the mass, m, and velocity, v) of the particle. i.e

λ ∝ $\frac{1}{mv}$

λ = $\frac{h}{mv}$            ------------------------(i)

Where;

h = constant of proportionality called Planck's constant = 6.626 x 10⁻³⁴Js

From the question;

λ = 0.950fm = 0.950 x 10⁻¹⁵m

v = 241ms⁻¹

Substitute these values into equation (i) as follows;

0.950 x 10⁻¹⁵ = $\frac{6.626 * 10^{-34}}{m*241}$

Solve for m;

m = $\frac{6.626 * 10^{-34}}{0.950*10^{-15} * 241}$

m = 0.0289 x 10⁻¹⁹kg

Therefore, the maximum mass of the object is 0.0289 x 10⁻¹⁹kg