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An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will be
- a)8.16 g/cm3
- b)6.81 g/cm3
- c)68.1 g/cm3
- d)1.86 g/cm3
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
An element with bcc geometry has atomic mass 50 and edge length 290 pm...
Length of the edge , a = 290 pm =290 x 10-10 cm
Volume of unit cell = ( 290 x 10-10 cm )3 = 24.39 x 10-24 cm3
Since it is bcc arrangement,
Number of atoms in the unit cell, Z = 2
Atomic mass of the element = 50
Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 1023
Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 1023 = 100/6.23 x 1023
Therefore , density = mass of unit cell / volume of unit cell
= 100/6.023 x 1023 x 24.39 x 10-24 = 6.81 g cm-3
Volume of unit cell = ( 290 x 10-10 cm )3 = 24.39 x 10-24 cm3
Since it is bcc arrangement,
Number of atoms in the unit cell, Z = 2
Atomic mass of the element = 50
Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 1023
Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 1023 = 100/6.23 x 1023
Therefore , density = mass of unit cell / volume of unit cell
= 100/6.023 x 1023 x 24.39 x 10-24 = 6.81 g cm-3
Most Upvoted Answer
An element with bcc geometry has atomic mass 50 and edge length 290 pm...
Length of the edge , a = 290 pm =290 x 10^-10 cm
Volume of unit cell = ( 290 x 10^-10 cm )^3 = 24.39 x 10^-24 cm^3
Since it is bcc arrangement,
Number of atoms in the unit cell, Z = 2
Atomic mass of the element = 50
Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 10^23
Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 10^23 = 100/6.23 x 10^23
Therefore , density = mass of unit cell / volume of unit cell
= 100/6.023 x 10^23 x 24.39 x 10^-24 = 6.81 g cm^-3
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Community Answer
An element with bcc geometry has atomic mass 50 and edge length 290 pm...
Density = z. M/a ^n(not) Where z =2 for bcc M=50 A=290 pm N(not) =6.022*10^23 On solving ur answer comes out to be 6.808 that is equivalent to 6.81g/cm^3
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An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer?.
An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will bea)8.16 g/cm3b)6.81 g/cm3c)68.1 g/cm3d)1.86 g/cm3Correct answer is option 'B'. Can you explain this answer?.
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