Question

Assuming that the smallest measurable wavelength in an experiment is 0.530 fm, what is the maximum mass of an object traveling at 563 m-s-for which the de Broglie wavelength is observable? 2.22 x10 kg

Answer 1

Given,

debroglie Wavelength (lambda)= 0.530 fm=(0.530 *10^-15) m

V=563m/s

To calculate the mass of the particle we will use the debroglie Wavelength equation-

Lambda=h/ mv

Where h is Planck's constant

m is mass of the the particle and v with which the particle moves.

So by putting the given values in the equation, we will get mass of the particle(m)

See the following image.

d h mr m = h Lhore h = 6.626 X 10 dr 6.626 X 10 34 Tis 0.5308 10 15m x 563 mis m = 6.6268 10-24 +15 0.53XSS? 6.626810 -19 J2 298.39 22.2 x 10 -22 J2 Js2 22 12 22 This can be unit of mass E = mc2 E C2 Unit 212 CS Scanned with CamScanner