Question

..An element has cube type unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of the unit cell is 24 × 10-²⁴ cm³ and density of the element 7.2 gram/cm³. Calculate the atoms present in 100 gram of the element.​

Answer 1

Question:

An element has cube type unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of the unit cell is 24 × 10-²⁴ cm³ and density of the element 7.2 gram/cm³. Calculate the atoms present in 100 gram of the element.

Given:

Density =7.2g/cm³

Volume= $24 × 10^{-24}cm^3$

Solution:

No of atoms in 1 unit cube= no. of atoms on one of its diagonals + no. of atoms on each corner.

$=> 2+ \frac{1}{8} \times 8 =>2+1=3$

We know $Density = \frac{Mass} {Volume}$

$7.2g/cm^3=\frac{Mass} {24 × 10^{-24}cm^3} \\\\ 7.2 \times 24 × 10^{-24}= Mass \\\\ Mass=172.8 \times 10^{-24} \\\\Mass= 1.728 \times 10^{-22}g$

This is the calculated mass of one unit of atom and we have calculated above that each unit has 3 atoms in total.

3 atoms has mass $1.728 \times 10^{-22}g$

$1 ~atom~ has ~mass ~\frac {1.728 \times 10^{-22}} {3}g \\\\ => 0.576 \times10^{-22}g \\\\ =>5.76 \times 10^{-23}g$

Now, $5.76 \times 10^{-23}g$ contains one atom.

As per the question, 100g will contain how many no. of atom

$\frac{1}{5.76 \times 10^{-23}g}\times 100g \\\\ \frac{100 \times 10^{23}}{5.76} \\\\ 17.36 \times10^{23} \\\\ 1.736 \times 10^{22} atoms$

Answer 2

Answer:

Sorry dear✌✌

Given:

Density =7.2g/cm³

Volume= 24 × 10^{-24}cm^324×10−24cm3

Solution:

No of atoms in 1 unit cube= no. of atoms on one of its diagonals + no. of atoms on each corner.

= > 2+ \frac{1}{8} \times 8 = > 2+1=3=>2+81×8=>2+1=3

We know Density = \frac{Mass} {Volume}Density=VolumeMass

\begin{gathered} 7.2g/cm^3=\frac{Mass} {24 × 10^{-24}cm^3} \\\\ 7.2 \times 24 × 10^{-24}= Mass \\\\ Mass=172.8 \times 10^{-24} \\\\Mass= 1.728 \times 10^{-22}g\end{gathered}7.2g/cm3=24×10−24cm3Mass7.2×24×10−24=MassMass=172.8×10−24Mass=1.728×10−22g

Gɪᴠᴇɴ :

The nᵗʰ term of an arithmetic sequence is 4.

The common difference (d) is - 4

The number of terms, n = 7

Tᴏ Fɪɴᴅ :

The first term (a) of the given sequence.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}★an=a+(n−1)d

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

↝ nᵗʰ term of an arithmetic sequence is 4

So,

Using formula

↝ nᵗʰ term of an arithmetic sequence is,

{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}an=a+(n−1)d

\rm :\implies\:4 = a + (7 - 1) \times ( - 4):⟹4=a+(7−1)×(−4)

\rm :\implies\:4 = a + 6 \times ( - 4):⟹4=a+6×(−4)

\rm :\implies\:4 = a - 24:⟹4=a−24

\begin{gathered} \longmapsto \: \begin{gathered}\:{\underline{\boxed{\bf{\blue{{\tt \: \: a \: = \: \: 28 }}}}}} \\ \end{gathered}\end{gathered}⟼a=28

Hence,

first term of the sequence is 28.