Question

A) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

B) Using the Bohr’s model calculate the orbital period in the 𝑛 = 1, 2, and 3 levels. (Speed of electron in 𝑛=1,2 and 3 level is $2.18\times {10}^{6}m/s;1.09\times {10}^{6}m/s;7.27\times {10}^{5}m/s$ respectively)
$=6.6\times {10}^{-34}Js$
$T_3=\frac{1}{(7.27\times {10}^5)}$
$\left(\frac{\left(3{)}^2\right(6.6\times {10}^{-34}{)}^2}{2\times 3.14(9.1\times {10}^{-31})(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}\right)$
$T-3=4.11\times {10}^{-15}s$
Final answer : $1.52\times {10}^{-16}s;$
$1.22\times {10}^{-15}s;4.11\times {10}^{-15}s$

Answer 1

Step 1: Estimate the expression for orbital speed of electron using Bohr’s model.
Formula used : $\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$
According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +𝑒. Then the centripetal force is provided by the Coulomb force. So,
$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$
$m{v}^2=\frac{k{e}^2}{r}.....\left(1\right)$
From Bohr’s quantization rule
$mvr=\frac{nh}{2\pi }$
$r=\frac{nh}{2\pi mv}.....\left(2\right)$
Use this value in equation (1)
$m{v}^2=\frac{\frac{k{e}^2}{nh}}{2\pi mv}$
$v=\frac{k{e}^2}{n\left(\frac{h}{2\pi }\right)}$
Step 2: Calculate speed of electron in H-atom in n = 1 level.
For n = 1,
$v_1=\frac{k{e}^2}{\frac{h}{2\pi }}$
Here, $k=9\times {10}^{9}Nm62/C^2$
e = charge on electron $=1.6\times {10}^{-19}C$
h = Plnak's constant $=6.6\times {10}^{-34}Js$
$v_1=\frac{(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}{\frac{6.6\times {10}^{-34}}{2\times 3.14}}$
$v_1=2.18\times {10}^{6}m/s$
Step 3: Calculate speed of electron in H-atom in n =2 level.
For n = 2,
$v_2=\frac{k{e}^2}{2\left(\frac{h}{2\pi }\right)}$
Here, $k=9\times {10}^{9}N{m}^2/C^2$
e = charge on electron $=1.6\times {10}^{-19}C$
h = Plank's constant $=6.6\times {10}^{-34}Js$
$v_2=\frac{(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}{\frac{6.6\times {10}^{-34}}{3.14}}$
$v_2=1.09\times {10}^{6}m/s$
Step 4: Calculate speed of electron in H-atom in n = 3 level.
For $n=3,v_3=\frac{k{e}^2}{3\left(\frac{h}{2\pi }\right)}$
Here, $k=9\times {10}^{9}N{m}^2/C^2$
e = charge on electron $=1.6\times {10}^{-19}C$
h = Plank's constant $=6.6\times {10}^{-34}Js$
$v_3=\frac{(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}{3\left(\frac{6.6\times {10}^{-34}}{2\times 3.14}\right)}$
$v_3=7.27\times {10}^{5}m/s$
Final answer : $2.18\times {10}^{6}m/s;1.09\times {10}^{6}m/s;$

B) Step 1: Estimate the expression for orbital period of electron using Bohr’s model.
Orbital period, $T=\frac{2\pi r}{v}$
According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +𝑒. Then the centripetal force is provided by the Coulomb force. So,
$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$
$m{v}^2=\frac{k{e}^2}{r}.....\left(2\right)$
From Bohr’s quantization rule,
$mvr=\frac{nh}{2\pi }.....\left(3\right)$
By squaring equation (3) and divide by equation (2)
$m{r}^2=\frac{\frac{n^2{h}^2}{4{\pi }^{2}k{e}^2}}{r}$
$r=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}$
Substitute this value in equation (1)
$T=\frac{2\pi }{v}\left(\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}\right)$
$T=\frac{1}{v}\left(\frac{n^2{h}^2}{2\pi mk{e}^2}\right)$
Step 2: Calculate orbital period of electron in H-atom in n = 1 level.
Given, speed of electron in n = 1 level, $v_1=2.18\times {10}^{6}m/s$
For n = 1,
$T_1=\frac{1}{v_1}\left(\frac{(1{)}^2{h}^2}{2\pi mk{e}^2}\right)$
Here, $k=9\times {10}^{9}N{m}^2/C^2$
e = charge on electron = $1.6\times {10}^{-19}C$
m = mass of electron $=9.1\times {10}^{-31}\text{kg}$
h = Plank′s constant $=.6.6\times {10}^{-34}Js$
$T_1=\frac{1}{(2.18\times {10}^6}$
$\left(\frac{\left(1{)}^2\right(6.6\times {10}^{-34}{)}^2}{2\times 3.14(91.\times {10}^{-31})(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}\right)$
$T_1=1.52\times {10}^{-16}S$
Step 3: Calculate orbital period of electron in H-atom in n = 2 level.
Given, speed of electron in n = 2 level,
$v_2=1.09\times {10}^{6}m/s$
Fo rn = 2,
$T_2=\frac{1}{v_2}\left(\frac{(2{)}^2{h}^2}{2\pi mk{e}^2}\right)$
Here, $k=9\times {10}^{9}N{m}^2/C^2$
e = charge on electron $=1.6\times {10}^{-19}C$
m = mass of electron $=9.1\times {10}^{-31}\text{kg}$
h = Plnak's constant $=6.6\times {10}^{-34}Js$
$T_2=\frac{1}{(1.09\times {10}^6}$
$\left(\frac{\left(2{)}^2\right(6.6\times {10}^{-34}{)}^2}{2\times 3.14(9.1\times {10}^{-31})(9\times {10}^9)(1.6\times {10}^{-19}{)}^2}\right)$
$T_2=1.22\times {10}^{-15}s$
Step 4: Calculate orbital period of electron in H-atom in n = 3 level.
Given, speed of electron in n = 3 level,
$v_3=7.27\times {10}^{5}m/s$
Fo rn = 3,
$T_3=\frac{1}{v_3}\left(\frac{(3{)}^2{h}^2}{2\pi mk{e}^2}\right)$
Here, $k=9\times {10}^{9}N{m}^2/C^2$
e = charge on electron
$=1.6\times {10}^{-19}C$
m = mass of electron
$=9.1\times {10}^{-31}\text{kg}$
h = Plank's constant