Question

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.

Answer 1

Answer:

The given atom is of Ca.

Explanation:

Given data:

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.