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

Step 1: Use the force balance in Bohr's model.
The Coulomb attraction supplies the centripetal force: $\dfrac{mv^2}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}$.
Step 2: Rearrange for the radius.
Cancelling one $r$ and solving gives $r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{mv^2}$.
Step 3: List the numbers.
Here $m = 9\times10^{-31}$ kg, $e = 1.6\times10^{-19}$ C, $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9$, and $v^2 = 25.6\times10^{10}$ m$^2$/s$^2$.
Step 4: Put the numbers in.
\[ r = \frac{(9\times10^9)(1.6\times10^{-19})^2}{(9\times10^{-31})(25.6\times10^{10})} = \frac{23.04\times10^{-29}}{230.4\times10^{-21}} \]
Step 5: Simplify the arithmetic.
The ratio gives $r = 0.1\times10^{-8}$ m for this excited orbit.
Step 6: Read off x.
Comparing the excited-state radius with the listed options, the matching value is $x = 4$, that is option B.
\[ \boxed{x = 4} \]