Q131 / 180 · NEET 2026 (Re-NEET) · Physics

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6}\times10^5$ ms$^{-1}$. The radius of the orbit is $x\times10^{-9}$ m. The value of $x$ is :
[Take the mass of electron to be $9\times10^{-31}$ kg, charge of electron $=-1.6\times10^{-19}$ C and $\dfrac{1}{4\pi\epsilon_0}=9\times10^9$ N m$^2$ C$^{-2}$]

  • Correct answer
    A
    $1$
  • B
    $4$
  • C
    $3$
  • D
    $2$

Solution

The electrostatic force between the nucleus and the electron provides the necessary centripetal force for the electron to revolve in its orbit.

$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}$

Rearranging for the radius $r$:

$r = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{mv^2}$

Substituting the given values:
$\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m$^2$ C$^{-2}$
$e = 1.6 \times 10^{-19}$ C
$m = 9 \times 10^{-31}$ kg
$v = \sqrt{25.6} \times 10^5$ m s$^{-1}$

$r = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times (\sqrt{25.6} \times 10^5)^2}$

$r = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-31} \times 25.6 \times 10^{10}}$

$r = \dfrac{2.56 \times 10^{-29}}{25.6 \times 10^{-21}}$

$r = 0.1 \times 10^{-8}$ m

$r = 1 \times 10^{-9}$ m

Comparing this with $r = x \times 10^{-9}$ m, we get:

$x = 1$

Answer: $1$

Hence, the correct answer is option: 'A'.

Relevant Questions