Question

the unit cell of Na is bcc and it's density is 0.97 g/cm3 what is radius of a sodium atom if the molar mass of Na is 23 g/mol​

Answer 1

Answer:

1.86×10-8 cm

Explanation:

Answer 2

Answer:

The radius of a sodium atom is approximately 1.86 Å.

Explanation:

To find the radius of a sodium atom, we can use the following formula:

density = (n x molar mass) / (V x N_A)

where n is the number of atoms in a unit cell, V is the volume of a unit cell, and N_A is Avogadro's constant.

For bcc unit cell, the number of atoms in a unit cell (n) is 2, and the volume of a unit cell (V) is given by:

V = (4r / √3)^3, where r is the radius of a sodium atom.

Substituting the given values and simplifying, we get:

0.97 g/cm^3 = (2 x 23 g/mol) / [(4r / √3)^3 x 6.022 x 10^23 /mol]

Solving for r, we get:

r = [(3 / 4) x (2 x 23 g/mol) / (0.97 g/cm^3 x 6.022 x 10^23 /mol)]^(1/3)

r = 1.86 Å (approx)

Therefore, the radius of a sodium atom is approximately 1.86 Å.

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