Self Studies

Solid State Test - 60

Result Self Studies

Solid State Test - 60
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    A metallic crystal crystallizes into a lattice containing a sequence of layers $$ABABAB$$.... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ?
  • Question 2
    1 / -0
    Calculate the density (in g cm$$^{-3}$$) of diamond from the fact that it has face-centered cubic structure with two atoms per lattice point and unit cell edge length of $$3.569\, \mathring A$$.
    Solution
    $$density = \dfrac{zM}{N_A \times (a)^3} $$

    For FCC, z=2

    $$density = \dfrac{(2 \times 4) \times 12}{6 \times 10^{23} \times (3.569 \times 10^{-8})^3} = 3.5 \text{ g cm}^{-3} $$
  • Question 3
    1 / -0
    A metal crystallizes in bcc lattice. The percent fraction of edge length not covered by atom is :
    Solution
    In BCC lattice, $$ 4r = a\sqrt{3}$$
    Fraction of each edge covered by atoms is $$ \dfrac{2r}{a} = \dfrac{a\sqrt{3}}{2a} = 86.6 \% $$
    Therefore the fraction of edge not covered by atom is $$ 13.4 \% $$
  • Question 4
    1 / -0
    A metal crystallizes into two cubic phases, face-centered cubic and body-centered cubic which have unit cell lengths as $$3.5\ \mathring A$$ and $$3.0\ \mathring {A} $$, respectively. Calculate the ratio of densities of fcc and bcc.
    Solution
    As we know,
    Density $$(\rho )=\dfrac {Z_{eff}\times  Mw}{a^3\times N_A}$$

    For fcc, $$Z_{eff}=4$$
    $$\rho _{fcc}=\dfrac {4\times  Mw}{(3.5\times 10^{-8})^3 \times N_A}$$

    For bcc, $$Z_{eff}=2$$
    $$\rho _{fcc}=\dfrac {2\times  Mw}{(3\times 10^{-8})^3 \times N_A}$$

    $$\dfrac{\rho _{fcc}}{\rho_{bcc}}=\dfrac {4\times  (3)^3}{(3.5)^3 \times 2}=1.26$$
  • Question 5
    1 / -0

    The packing efficiency of a two-dimensional square unit cell shown is __________.

    Solution
    The diagonal is given as $$4r$$, where $$r$$ is the radius of an atom.

    Diagonal $$= \sqrt {L^2+L^2}=\sqrt {2L}$$

    $$4r$$ $$=\sqrt {2L}$$ or $$L =\dfrac {4r}{\sqrt 2}$$

    Total area $$=L^2$$

    or $$\left ( \dfrac {4r}{\sqrt 2} \right )^2=8r^2$$

    Number of spheres inside the square is

    $$1 + 4 \left ( \dfrac {1}{4} \right )=2$$

    Area of each sphere $$= \pi r^2$$

    Total area of spheres $$=2 \times \pi r^2$$

    Packing fraction $$=\dfrac {\text{Total area of spheres}}{\text{Total area}}$$

    $$= \dfrac{2 \times \pi r^2}{8 r^2}=\dfrac{\pi}{4}=0.785$$

    So, the percentage fraction is $$78.5\%$$.
  • Question 6
    1 / -0
    An element crystallizes in a structure having FCC unit cell of an edge length $$200$$ pm. Calculate the density (in g cm$$^{-3}$$) if $$200$$ g of this element contain $$5\, \times\, 10^{24}$$ atoms.
    Solution
    Molecular mass of the given element is $$ = \dfrac{200 \times 6 \times 10^{23}}{5 \times 10^{24}} = 24 $$
    $$density = \dfrac{zM}{N_A \times (a \times 10^{-10})^3} $$
    $$density = \dfrac{4 \times 24 }{6 \times 10^{23} \times (200 \times 10^{-10})^3} $$
    $$density = 20 \text{ g cm}^{-3} $$
  • Question 7
    1 / -0
    The $$\gamma$$-form of iron has fcc structure (edge length $$=386$$ pm) and $$\beta$$-form has bcc structure (edge length $$=290$$ pm). The ratio of density in $$\gamma$$-form to that in $$\beta$$-form is :
    Solution
    $$ \begin{array}{l} Z_{e f f} \text { for } \mathrm{fcc}=4 \text { and forbcc }=2 / \text { unit cell } \\ \dfrac{\rho_{\lambda}}{\rho_{\beta}}=\dfrac{\rho_{f c c}}{\rho_{f c c}}=\left(\dfrac{Z_{e f f(f c c)}}{Z_{e f f(b c c)}}\right) \times\left(\dfrac{a_{b c c}}{a_{f c c}}\right)^{3} \\ =\dfrac{4}{2} \times\left(\dfrac{290}{386}\right)^{3}=0.848 \end{array} $$
  • Question 8
    1 / -0
    If the lattice parameter of Si is $$5.43$$ $$\mathring {A} $$ and the mass of Si atom is $$28.08 \times 1.66 \times 10^{-27}$$ kg, the density of silicon in kg m$$^{-3}$$ is:
    [Given: Silicon has a diamond cubic structure.]
    Solution
    SInce Si has diamond cubic structure, its unit cell contains $$8$$ atoms.
    Density $$\displaystyle = \dfrac {\text{Mass}}{\text{Volume}} $$
    Mass $$\displaystyle = 8 \times 28.08 \times 1.66 \times 10^{-27} = 3.73  \times 10^{-25} $$ kg
    Volume $$\displaystyle  = (5.43 \times 10^{-10})^3 = 1.6 \times 10^{-28} \: $$ m$$^3$$
    Density $$\displaystyle = \frac {3.73  \times 10^{-25} }{ 1.6 \times 10^{-28}} = 2330$$ kg m$$^{-3}$$
  • Question 9
    1 / -0
    An element has a FCC structure with edge length 200 pm. Calculate density if 200 g of this element contains $$24 \times 10^{23}$$ atoms.
    Solution
    A fcc unit cell contains 4 atoms per unit cell.

    Mass of unit cell is equal to mass of 4 atoms. It is equal to $$\displaystyle 4 \times \frac {200}{24 \times 10^{23}} = 3.33 \times 10^{-22} $$ g

    Volume of the unit cell $$\displaystyle = (200 \times 10^{-10})^3  = 8 \times 10^{-24} \: cm^3$$

    $$\displaystyle Density = \frac {Mass}{Volume} =   \frac { 3.33 \times 10^{-22} }{8 \times 10^{-24}} = 41.6 \: g/cm^3$$
  • Question 10
    1 / -0

    Directions For Questions

    The number of Schottky defects $$\left(n\right)$$ present in an ionic containing $$N$$ ions at temperature $$T$$ is given by $$n = N{ e }^{ -{ E }/{ 2kT } }$$ where $$E$$ is energy required to create $$n$$ Schottky defects and $$k$$ is Boltzmann constant. The number of Frenkel defects $$\left(n\right)$$ in an ionic crystal having $$N$$ ions is given by $$n={ \left( \dfrac { N }{ { N }_{ i } }  \right)  }^{ \dfrac { 1 }{ 2 }  }{ e }^{ -{ E }/{ 2kT } }$$ where $$E$$ is energy required to create $$n$$ Frenkel defects and $${N}_{i}$$ is the number of interstitial sites.

    ...view full instructions

    Which of the following does not posses any plane of symmetry?
    Solution
    Triclinic crystal lattice does not posses any plane of symmetry.
    Plane of symmetry is that imaginary plane which passes through the centre of the crystal and divides it into two equal portions (just mirror images of each other).
    For triclinic crystal system, $$\displaystyle a \neq b \neq c, \alpha \neq \beta \neq \gamma = 90^o$$ 
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now