How many lithium atoms are present in a unit cell with edge length $3.5 ˚ A$ and density $0.53 g c m^{- 3}$ ? (Atomic mass of Li =6.94)

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Solution

The correct option is A 2
The number of Li atoms in a unit cell is 2.

The density of the atom is given by the expression

density=(Zm/N(a^3))

Here, Z is the number of atom in a unit cell, m is the molecular mass of the atom, N is the Avogadro's number, and a is edge length

Plugging the values in the above equation

0.53=

z=2

Therefore the number of atom in a unit cell of Li is 2.

Suggest Corrections

Similar questions

o A

^{- 3}

L i = 6.94

0.53 g / c m^{3}

L i

3.5 \circ A

6.94

0.53

^{- 3}

3.5 ˚ A

N_{A} = 6.02 \times 10^{23}, M = 6.94

^{- 1}

530 {kg m}^{- 3}

6.94 {g mol}^{- 1}

N_{A} = 6.02 \times 10^{23} {mol}^{- 1}

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