The speed of electron in the second orbit of `Be^(3+)` ion will be

Options

A. `( C)/(137)`
B. `( 2 C)/(137)`
C. `(3 C)/(137)`
D. `(4 C)/(137)`

Explanation

To find the speed of an electron in the second orbit of the \( Be^{3+} \) ion, we can use the formula for the speed of an electron in a hydrogen-like atom:

\[ v = 2.18 \times 10^6 \frac{Z}{n} \text{ m/s} \]

where:
- \( v \) is the speed of the electron,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 1: Identify the atomic number \( Z \) and principal quantum number \( n \)

For the \( Be^{3+} \) ion:
- The atomic number \( Z \) of beryllium (Be) is 4.
- Since we are looking for the speed in the second orbit, the principal quantum number \( n \) is 2.

### Step 2: Substitute the values into the formula

Now we can substitute \( Z \) and \( n \) into the formula:

\[ v = 2.18 \times 10^6 \frac{4}{2} \]

### Step 3: Simplify the expression

Calculating the fraction:

\[ \frac{4}{2} = 2 \]

Now substitute this back into the equation:

\[ v = 2.18 \times 10^6 \times 2 \]

### Step 4: Calculate the speed

Now perform the multiplication:

\[ v = 2.18 \times 10^6 \times 2 = 4.36 \times 10^6 \text{ m/s} \]

### Step 5: Express the speed in terms of the speed of light \( c \)

The speed of light \( c \) is given as \( 3 \times 10^8 \text{ m/s} \). To express \( v \) in terms of \( c \):

\[ \frac{v}{c} = \frac{4.36 \times 10^6}{3 \times 10^8} \]

### Step 6: Simplify the ratio

Calculating the ratio:

\[ \frac{v}{c} = \frac{4.36}{3} \times \frac{10^6}{10^8} = \frac{4.36}{3} \times 10^{-2} \]

This simplifies to:

\[ \frac{v}{c} \approx 0.1453 \]

### Step 7: Final expression

We can express \( v \) in terms of \( c \):

\[ v \approx \frac{2c}{137} \]

### Conclusion

Thus, the speed of the electron in the second orbit of the \( Be^{3+} \) ion is approximately \( \frac{2c}{137} \).

---