Question

In hydrogen atom, the electron is excited from ground state to higher energy state and its orbital velocity is reduced to $\dfrac{1}{{{3^{rd}}}}$ of its initial value. The radius of the orbit in the ground state is R. What is the radius of the orbit in that higher energy state is?
A) $2R$
B) $3R$
C) $27R$
D) $9R$

Answer 1

Hint: According to Bohr’s model for hydrogen atoms, the energy states are numbered as $1,2,3..$ which are called principal quantum numbers and radii of the orbits depends on these principal quantum numbers and velocity is inversely proportional to principal quantum number.

Complete answer
As we know that the stationary states lowest energy level of an atom is its ground state and when an electron in an atom absorbs energy it jumps to a higher state called excited state. Now according to Bohr’s model for hydrogen atom, he numbered the energy states as $1,2,3..$ which are commonly called principal quantum numbers.

He proved that the radii of the stationary orbit or ground state depends on the principal quantum number and the expression was given as: ${r_n} = {n^2}{a_0}$ and the velocity of electron depends inversely on the principal quantum number and the expression is given as: ${v_n} = \dfrac{1}{n}$.
So, from the given question we can make out that the velocity of the excited state is $\dfrac{1}{{{3^{rd}}}}$ of its ground state. Using the above formula and after comparing both we will get:
$\dfrac{{{v_e}}}{{{v_g}}} = \dfrac{1}{3}$, where ${v_e}$is the velocity of excited state and ${v_g}$ is the velocity of ground state, we will get $n = 3$.

Now, we know that radius is given as: ${r_n} = {n^2}{a_0}$ and we are given with the radius of ground state which is R , so we can calculate the radius of the excited state easily by:
$\dfrac{{{r_e}}}{{{r_g}}} = \dfrac{{{{(3)}^2}}}{{{{(1)}^2}}}$, where ${r_e}$is radius of excite state and ${r_g}$is ground state radius.
 $
  \dfrac{{{r_e}}}{R} = 9 \\
  {r_e} = 9R \\
 $

Therefore the correct answer is option (D).

Note:Remember that as the value of $n$ increases, the radius also increases or the distance between nucleus and electron increases and when the value of $n$ increases the velocity of electron decreases but as the positive charge on electron increases, velocity increases. The first orbit of a hydrogen atom is $52.9pm$ as it has only one electron in its first orbit and this orbit is called Bohr’s orbit.