Assuming that the smallest measurable wavelength in an experiment is 0.650 fm (femtometers), what is the maximum mass of an object traveling at 129 m·s–1 for which the de broglie wavelength is observable?

Hey there!:

Wavelength λ = 0.650 fm = 0.650* 10⁻¹⁵ m (1 fm = 10⁻¹⁵ m)

Speed v = 129 m/s⁻¹

Planck constant h = 6.626 * 10⁻³⁴ J.s

De Broglie equation: λ = h/mv

Mass m = h/λv

= 6.626 * 10⁻³⁴/ ( 0.650* 10⁻¹⁵ * 129)

= 6.626*10⁻³⁴ / 8.385*10⁻¹⁴

= 7.902*10⁻²¹ Kg

Hope that helps!

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.