Question

\(C_{11}H_{18}O_{12}\)The value of the Rydberg constant (\(R_H\)) is \(2.18 \times 10^{-18} \, \text{J}\). The velocity of an electron having mass \(9.1 \times 10^{-31} \, \text{kg}\) in Bohr's first orbit of the hydrogen atom is:\(\dots \times 10^5 \, \text{ms}^{-1}\) (nearest integer).

Answer 1

Correct Answer: 22

Using the formula for the velocity of an electron:

\( v = 2.18 \times 10^6 \times \frac{Z}{n} \)

For hydrogen (\( Z = 1, n = 1 \)):

\( v = 2.18 \times 10^6 \times \frac{1}{1} = 2.18 \times 10^6 \, \text{ms}^{-1} \approx 22 \times 10^5 \, \text{ms}^{-1} \)

Answer 2

Step 1: Recall the formula for the velocity of an electron in the nth Bohr orbit
In Bohr’s model for the hydrogen atom, the velocity of an electron in the nth orbit is given by:
\[ v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \] However, it can also be expressed in terms of the Rydberg constant \(R_H\) as:
\[ v_n = \sqrt{\frac{2R_H h}{m_e}} \cdot \frac{1}{n} \] But the simplest and direct Bohr-derived expression is:
\[ v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \] Yet since we are given \(R_H = 2.18 \times 10^{-18}\, \text{J}\), we can instead use the energy relation:
\[ E_n = -R_H \cdot \frac{1}{n^2} = \frac{1}{2} m_e v_n^2 \] for the first orbit (\(n = 1\)).

Step 2: Solve for \(v_1\)
\[ R_H = \frac{1}{2} m_e v_1^2 \] \[ v_1 = \sqrt{\frac{2R_H}{m_e}} \] Substitute the given values:
\[ R_H = 2.18 \times 10^{-18}\, \text{J}, \quad m_e = 9.1 \times 10^{-31}\, \text{kg}. \] \[ v_1 = \sqrt{\frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}}} \] \[ v_1 = \sqrt{\frac{4.36 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{4.79 \times 10^{12}} = 2.19 \times 10^{6}\, \text{m/s}. \]

Step 3: Express in the required form
\[ v_1 = 2.19 \times 10^{6}\, \text{m/s} = 21.9 \times 10^5\, \text{m/s}. \] Nearest integer value = 22 × 10⁵ m/s.

Step 4: Final answer

22