The number of unit cells present in 39 g of potassium if it crystallizes as body centred cube is (N= Avogadro number, At. wt. of potassium =39 )

Question

The number of unit cells present in 39 g of potassium if it crystallizes as body centred cube is (N= Avogadro number, At. wt. of potassium =39 ) (A) (N/4) (B) (N/2) (C) (N/3) (D) N

Tardigrade Admin · Accepted Answer

Potassium has bcc system Number of mole =39 / 39=1 mole 1 mole of atoms =6.022 × 1023 atoms =N ∵ In bcc system 2 atoms are present in 1 unit cell. ∴ N number of atoms are present in (N/2) unit

Q. The number of unit cells present in $39 g$ of potassium if it crystallizes as body centred cube is
$(N =$ Avogadro number, At. wt. of potassium $= 39$ )

$\frac{N}{4}$

$\frac{N}{2}$

$\frac{N}{3}$

$N$

Potassium has bcc system

Number of mole $= 39/39 = 1$ mole

$1$ mole of atoms $= 6.022 \times 1 0^{23}$ atoms $= N$

$∵$ In bcc system $2$ atoms are present in $1$ unit cell.

$∴ N$ number of atoms are present in $\frac{N}{2}$ unit

Q. The number of unit cells present in 39g of potassium if it crystallizes as body centred cube is (N= Avogadro number, At. wt. of potassium =39 )

4N​

2N​

3N​

N

Potassium has bcc system Number of mole =39/39=1 mole 1 mole of atoms =6.022×1023 atoms =N ∵ In bcc system 2 atoms are present in 1 unit cell. ∴N number of atoms are present in 2N​ unit

Q. The number of unit cells present in $39 g$ of potassium if it crystallizes as body centred cube is
$(N =$ Avogadro number, At. wt. of potassium $= 39$ )

$\frac{N}{4}$

$\frac{N}{2}$

$\frac{N}{3}$

$N$

Potassium has bcc system

Number of mole $= 39/39 = 1$ mole

$1$ mole of atoms $= 6.022 \times 1 0^{23}$ atoms $= N$

$∵$ In bcc system $2$ atoms are present in $1$ unit cell.

$∴ N$ number of atoms are present in $\frac{N}{2}$ unit