The equilibrium fraction of lattice sites that are vacant in silver (ag) at 500â°c is 0.5 ã 10-6. calculate the number of vacancies (per meter cubed) at 500â°c. assume a density of 10.35 g/cm3 for ag, and note that aag = 107.87 g/mol.

Answer : The number of vacancies (per meter cube) = 2.889 ×

Solution :  Given,

Atomic weight of silver (Ag), = 107.87 g/mol

Density of silver (Ag), = 10.35 = 10.35 × = 10.35 ×

Avogadro's number, = 6.022 × atoms/mol

Fraction of lattice sites that are vacant in silver = 0.5 ×

Formula used :

Where,

= Total number of lattice sites in Ag

= Avogadro's number

= Density of silver

= Atomic weight of silver

This problem is solved by two steps:

step 1 : First we calculate the total number of lattice sites in silver.

Now put all the values in above formula, we get

         = 5.778 × atoms/

Step 2 : Now multiply the value of by the fraction of lattice sites that are vacant in silver,  0.5 × , we get the number of vacancies (per meter cube)

The number of vacancies (per meter cube) = 5.778 × atoms/ × 0.5 × = Answer : The number of vacancies (per meter cube) = 2.889 ×

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explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.