The ratio of the volume of a tetragonal lattice unit cell to that of a hexagonal lattice unit cell is (both having same respective lengths)
a.`(sqrt3)/2abc`
b.`(2)/(3sqrt3)`
c.`(2)/(sqrt(3) (a^(2)c)/(b)`
d.`1`

Options

A. `(sqrt3)/2abc`
B. `(2)/(3sqrt3)`
C. `(2)/(sqrt(3) (a^(2)c)/(b)`
D. `1`

Explanation

To find the ratio of the volume of a tetragonal lattice unit cell to that of a hexagonal lattice unit cell, we start by calculating the volumes of both unit cells.

### Step 1: Calculate the volume of the hexagonal unit cell
The volume \( V_h \) of a hexagonal unit cell is given by the formula:

\[ V_h = \frac{3\sqrt{3}}{2} a^2 c \]

where \( a \) is the length of the base and \( c \) is the height.

### Step 2: Calculate the volume of the tetragonal unit cell
The volume \( V_t \) of a tetragonal unit cell is given by the formula:

\[ V_t = a^2 c \]

where \( a \) is the length of the sides and \( c \) is the height.

### Step 3: Set up the ratio of the volumes
Now we need to find the ratio of the volume of the tetragonal unit cell to that of the hexagonal unit cell:

\[ \text{Ratio} = \frac{V_t}{V_h} = \frac{a^2 c}{\frac{3\sqrt{3}}{2} a^2 c} \]

### Step 4: Simplify the ratio
In this ratio, \( a^2 \) and \( c \) in the numerator and denominator will cancel out:

\[ \text{Ratio} = \frac{1}{\frac{3\sqrt{3}}{2}} = \frac{2}{3\sqrt{3}} \]

### Conclusion
Thus, the ratio of the volume of a tetragonal lattice unit cell to that of a hexagonal lattice unit cell is:

\[ \frac{2}{3\sqrt{3}} \]

### Final Answer
The correct answer is option **b. \( \frac{2}{3\sqrt{3}} \)**.

---