Question

The velocity of an electron in the seventh orbit of hydrogen-like atom is 3.6 × 10⁶ m/s. Find the velocity of the electron in the 3^rd orbit.

• A. 4.2 × 10⁶ m/s
• B. 8.4 × 10⁶ m/s
• C. 2.1 × 10⁶ m/s
• D. 3.6 × 10⁶ m/s

Answer 1

The Correct Option is B

 To find the velocity of the electron in the third orbit of a hydrogen-like atom, we can use the formula for the velocity of an electron in the \(n\)^th orbit of a hydrogen-like atom:

\(v_n = \dfrac{v_0}{n}\)

where \(v_0\) is the velocity of the electron in the first orbit. For hydrogen, this is approximately \(2.18 \times 10^6 \, \text{m/s}\). For hydrogen-like atoms, the formula requires adjustment for both neutron number (Z) and orbit number (n):

\(v_n = \dfrac{2.18 \times 10^6 \, \text{m/s} \times Z}{n}\)

In this problem, we are given:

• The velocity of the electron in the seventh orbit (\(v_7 = 3.6 \times 10^6 \, \text{m/s}\))

Using the equation for the 7th orbit, we have:

\(v_7 = \dfrac{2.18 \times 10^6 \times Z}{7} = 3.6 \times 10^6\)

From this, we can determine \(Z\):

\(2.18 \times 10^6 \times Z = 3.6 \times 10^6 \times 7\)

\(Z = \dfrac{3.6 \times 10^6 \times 7}{2.18 \times 10^6}\)

Solving for \(Z\), we find:

\(Z \approx 11.56\)

We will assume \(Z \approx 12\) (for a hypothetical hydrogen-like element).

Next, we calculate the velocity of the electron in the third orbit:

\(v_3 = \dfrac{2.18 \times 10^6 \times 12}{3}\)

\(= 8.4 \times 10^6 \, \text{m/s}\)

Thus, the velocity of the electron in the third orbit is 8.4 × 10⁶ m/s.

The correct answer is

8.4 × 10⁶ m/s

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