An atom of a particular element is traveling at 1.00% of the speed of light. The de Broglie wavelength is found to be 3.31 * 10-3 pm. Which element is this? Prove it.

The given atom is of Ca.

Explanation:

Given dа ta:

Speed of atom = 1% of speed of light

De-broglie wavelength = 3.31*10⁻³ pm (3.31*10⁻³ / 10¹² = 3.31*10⁻¹⁵ m)

What is element = ?

Solution:

Formula:

m = h/λv

m = mass of particle

h = planks constant

v = speed of particle

λ = wavelength

Now we will put the values in formula.

m = h/λv

m = 6.63*10⁻³⁴kg. m². s⁻¹/3.31*10⁻¹⁵ m * (1/100) * 3*10⁸ m/s

m = 6.63*10⁻³⁴kg. m². s⁻¹ / 0.099*10⁻⁷m²/s

m = 66.97*10⁻²⁷ Kg/atom

or

6.69*10⁻²⁶ Kg/atom

Now here we will use the Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms, ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 * 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 * 10²³ molecules of water

Now in given problem,

6.69*10⁻²⁶ Kg/atom * 6.022 * 10²³ atoms / mol * 1000 g / 1kg

40.3*10⁻³*10³g/mol

40.3 g/mol

So the given atom is of Ca.