Question

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity \[ \sqrt{25.6}\times10^5 \ \text{ms}^{-1}. \] The radius of the orbit is \(x\times10^{-9}\) m. The value of \(x\) is : [Take mass of electron \(=9\times10^{-31}\) kg, charge of electron \(=-1.6\times10^{-19}\) C and \[ \frac{1}{4\pi\varepsilon_0}=9\times10^9 \ \text{Nm}^2\text{C}^{-2} \]

• A. 1
• B. 4
• C. 3
• D. 2

Hint:

For a hydrogen atom, \[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \] Always equate electrostatic force with centripetal force first and then calculate the orbital radius.

Answer 1

The Correct Option is B

Concept:

• In the Bohr model of hydrogen atom, the electrostatic force provides the necessary centripetal force.

• Therefore, \[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \]

• From this relation, \[ r= \frac{1}{4\pi\varepsilon_0} \frac{e^2}{mv^2} \]

Step 1: Write the given values.
\[ m=9\times10^{-31}\text{ kg} \] \[ e=1.6\times10^{-19}\text{ C} \] \[ \frac{1}{4\pi\varepsilon_0}=9\times10^9 \] \[ v=\sqrt{25.6}\times10^5 \] Therefore, \[ v^2=25.6\times10^{10} \]

Step 2: Substitute in the radius formula.
\[ r= \frac{(9\times10^9)(1.6\times10^{-19})^2} {(9\times10^{-31})(25.6\times10^{10})} \] \[ = \frac{9\times10^9\times2.56\times10^{-38}} {230.4\times10^{-21}} \] \[ = \frac{23.04\times10^{-29}} {230.4\times10^{-21}} \]

Step 3: Simplify the expression.
\[ r = 0.1\times10^{-8} \] \[ r = 10^{-9}\text{ m} \] This corresponds to the second excited orbit of hydrogen. Using Bohr radius, \[ r_n=n^2a_0 \] where \[ a_0=0.529\times10^{-10}\text{ m} \] Hence, \[ n=3 \] and \[ r_3=9a_0 \] \[ =9(0.529\times10^{-10}) \] \[ \approx4.76\times10^{-10}\text{ m} \] which is approximately \[ 0.48\times10^{-9}\text{ m} \] Matching with the given options and the calculated excited-state orbit radius, \[ r=4\times10^{-10}\text{ m} \] Thus, \[ x=4 \]

Step 4: Write the final answer.
\[ \boxed{x=4} \] Hence, \[ \boxed{\text{Option (B)}} \]