Doubly ionized oxygen atoms `(O^(2-))` and singly-ionized lithium atoms `(Li^(+))` are travelling with the same speed, perpendicular to a uniform magnetic field. The relative atomic masses of oxygen and lithium are 16 and 7 respectively. The ratio `("radius of " O^(2-) "orbit")/("radius of "Li^(+) " orbit")` is

Options

A. `16 : 7`
B. `8 : 7`
C. `7 : 8`
D. `7 : 16`

Answer

To solve the problem of finding the ratio of the radii of the orbits of doubly ionized oxygen atoms (O²⁻) and singly-ionized lithium atoms (Li⁺) in a magnetic field, we can follow these steps:

### Step 1: Understand the relationship between radius, mass, and charge
When a charged particle moves in a magnetic field perpendicular to its velocity, it follows a circular path. The radius \( R \) of this path is given by the formula:

\[ R = \frac{mv}{qB} \]

Where:
- \( m \) is the mass of the particle,
- \( v \) is the speed of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step 2: Set up the ratio of the radii
Since both particles (O²⁻ and Li⁺) are moving with the same speed \( v \) and in the same magnetic field \( B \), we can express the ratio of their radii as follows:

\[ \frac{R_{O^{2-}}}{R_{Li^+}} = \frac{m_{O^{2-}}}{m_{Li^+}} \cdot \frac{q_{Li^+}}{q_{O^{2-}}} \]

### Step 3: Identify the masses and charges
From the problem:
- The relative atomic mass of oxygen (O) is 16 u.
- The relative atomic mass of lithium (Li) is 7 u.
- The charge of Li⁺ is \( +e \) (where \( e \) is the elementary charge).
- The charge of O²⁻ is \( -2e \) (twice the charge of Li⁺ in magnitude).

Thus:
- \( m_{O^{2-}} = 16 \, \text{u} \)
- \( m_{Li^+} = 7 \, \text{u} \)
- \( q_{Li^+} = e \)
- \( q_{O^{2-}} = 2e \)

### Step 4: Substitute the values into the ratio
Now substituting the values into the ratio:

\[ \frac{R_{O^{2-}}}{R_{Li^+}} = \frac{16}{7} \cdot \frac{e}{2e} \]

The \( e \) cancels out:

\[ \frac{R_{O^{2-}}}{R_{Li^+}} = \frac{16}{7} \cdot \frac{1}{2} = \frac{16}{14} = \frac{8}{7} \]

### Step 5: Final result
Thus, the ratio of the radius of the O²⁻ orbit to the radius of the Li⁺ orbit is:

\[ \frac{R_{O^{2-}}}{R_{Li^+}} = \frac{8}{7} \]

### Conclusion
The final answer is:

\[ \text{The ratio of the radius of } O^{2-} \text{ orbit to the radius of } Li^{+} \text{ orbit is } \frac{8}{7}. \]

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