Question

An electron Jumps from the first excited state to the ground state of the hydrogen atom. What will be the percentage change in the speed of the electron?
A. 25%
B. 50%
C. 100%
D. 200%

Answer 1

-Hint: In this question we will proceed with the formula which is obtained on the basis of Bohr’s postulates which shows that the velocity of electrons is in inverse relation with the number of orbits.
Then after we will find the ratio to get the desired result.
Using Formula: -
According to Bohr’s postulate of atomic model we know that the velocity of an electron revolving along the nucleus of an atom is given by
$v = \dfrac{{2\pi kZ{e^2}}}{{nh}}$
Where $v,\,k,\,Z,\,n\,and\,h$ are the velocity of electron, electrostatic constant, atomic number of atoms, number of orbit and Planck’s constant respectively.
Given Data: -
For hydrogen atom
${n_1} = 1$
${n_2} = 2$
$Z = 1\,\,$

Step by step solution:
For electron orbiting in first orbit the velocity of electron given by
${v_1} = \dfrac{{2\pi k{e^2}}}{{{n_1}h}}\,.......................(1)$
For electron orbiting in second orbit the velocity of electron given by
${v_2} = \dfrac{{2\pi k{e^2}}}{{{n_2}h}}\,......................(2)$
On dividing $eq.\,(2)\,by\,eq.\,(1)$we will get as follows:
$\dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{{n_1}}}{{{n_2}}}$
On subtracting the above result by $1$ on both sides, we will get as follows
$1 - \dfrac{{{v_2}}}{{{v_1}}} = 1 - \dfrac{{{n_1}}}{{{n_2}}}$
$\dfrac{{{v_1} - {v_2}}}{{{v_1}}} = \dfrac{{\Delta v}}{{{v_1}}} = \dfrac{{{n_2} - {n_1}}}{{{n_2}}} = \dfrac{{2 - 1}}{2} = \dfrac{1}{2}$
$\Delta v\,\% = \dfrac{{\Delta v}}{{{v_1}}} \times 100 = \dfrac{1}{2} \times 100 = 50\,\%$

Hence the option $(B)$is correct.

Note: - In this question we should be aware of the Bohr’s postulates for the atomic model and the relation between velocity of an electron orbiting the nucleus and number of orbits of the atom. We should also be known about the basic mathematical operations to make the ratio of any quantity into the percentage change of that quantity.