\(C_{11}H_{18}O_{12}\)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).
Correct Answer: 22
Approach Solution - 1
Using the formula for the velocity of an electron:
\( v = 2.18 \times 10^6 \times \frac{Z}{n} \)
For hydrogen (\( Z = 1, n = 1 \)):
\( v = 2.18 \times 10^6 \times \frac{1}{1} = 2.18 \times 10^6 \, \text{ms}^{-1} \approx 22 \times 10^5 \, \text{ms}^{-1} \)
Approach Solution -2
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
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