Question

Assuming that the smallest measurable wavelength in an experiment is 0.430 fm , what is the maximum mass of an object traveling at 885 m⋅s−1 for which the de Broglie wavelength is observable?

Answer 1

Answer:- The maximum mass of the object is
`1.74*10^-^2^1kg` .

Solution:- The equation for de-Broglie wavelength is:

`\lambda =(h)/(mv)`

where, h is planck's constant, m is the mass and v is the velocity.

Given wavelength is 0.430 fm. fm stands for femtometer. let's convert it to meter.

`1fm=10^-^1^5m`

`0.430fm((10^-^1^5m)/(1fm))`

=
`4.30*10^-^1^6m`

velocity is given as
`885m.s^-^1` . Let's plug in the values in the equation:

`4.30*10^-^1^6m=(6.626*10^-^3^4J.s)/(m*885m.s^-^1)`

(at denominator the first m stands for mass and the second m is the unit of length that stands for meter)

We know that,
`J=kg.m^2.s^-^2`

So,
`4.30*10^-^1^6m=(6.626*10^-^3^4kg.m^2.s^-^1)/(m*885m.s^-^1)`

`4.30*10^-^1^6m=(7.48*10^-^3^7kg.m)/(m)`

`m=(7.48*10^-^3^7kg.m)/(4.30*10^-^1^6m)`

`m=1.74*10^-^2^1kg`

So, the maximum mass of the object is
`1.74*10^-^2^1kg` .