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) Calculate the orbital period in each of these levels.

Answer 1

(a) Let \(v_1\) be the orbital speed of the electron in a hydrogen atom in the ground state level, \(n_1=1\). For charge (e) of an electron, \(v_1\) is given by the relation,
\(v_1=\frac{e^2}{n_14π∈_0(\frac{h}{2π})}=\frac{e^2}{2∈_0h}\)
Where, \(e=1.6×10^{-19}C\)
\(∈_0\)=Permittivity of free space\(=8.85×10^{-12}N^{-1}C^2m^{-2}\)
h=planck's constant\(=6.62×10^{-34}Js\)
\(∴v_1=\frac{(1.6×10^{-19})^2}{2×8.85×10^{-12}×6.62×10^{-34}}\)
\(=0.0218×10^8=2.18×10^6m/s\)
For level \(n_2=2\), we can write the relation for thecorresponding orbital speed as:
\(v_2=\frac{e^2}{n^2_2∈_0h}\)
\(=\frac{(1.6×10^{-19})^2}{3×2×8.85×10^{-12}×6.62×10^{-34}}\)
\(=1.09×10^6m/s\)
And, for \(n_3=3\), we can write the relation for the corresponding orbital speed as:
\(v_3=\frac{e^2}{n_32∈_0h}\)
\(=\frac{(1.6×10^{-19})^2}{3×2×8.85×10^{-12}×6.62×10^{-34}}\)
\(=7.27×10^5m/s\)
Hence,the speed of the electron in a hydrogen atom in n=1, n=2, and n=3 is \(2.18×10^6m/s,1.09×10^6m/s,7.27×10^5m/s\) respectively.

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(b)Let \(T_1\) be the orbital peroid of the electron when it is in level \(n_1=1\).
Orbital period is related to orbital speed as:
\(T_1=\frac{2πr_1}{v_1}\)
Where, \(r_1\)=Radius of the orbit\(=\frac{n_1^2h^2∈_0}{πme^2}\)
h=planck's constant\(=6.62×10^{-34}Js\)
e=charge on an electron\(=1.6×10^{-19}C\)
\(∈_0=\)permittivity of free space\(=8.85×10^{-12}N^{-1}C^2m^{-2}\)
m=Mass of an electron\(=9.1×10^{-31}kg\)
\(∴T_1=\frac{2πr_1}{v_1}\)
\(=\frac{2π×(1)^2×(6.62×10^{-34})^2×8.85×10^{-12}}{2.18×10^6×π \times 9.1×10^{-31}×(1.6×10^{-19})^2}\)
\(=15.27×10^{-17}=1.527×10^{-16}s\)
For level \(n_2=2\), we can write the period as:
\(T_2=\frac{2πr_2}{v_2}\)
Where \(r_2\)=Radius of the electron in \(n_2=2\)
\(=\frac{(n_2)^2h^2∈_0}{πme^2}\)
\(∴T_2=\frac{2πr_2}{v_2}\)
\(=\frac{2π×(2)^2×(6.62×10^{-34})^2×8.85×10^{-12}}{1.09×10^6×π×9.1×10^{-31}×(1.6×10^{-19})^2}\)
\(=1.22×10^{-15}s\)
And, for level \(n_3=3\), we can write the period as:
\(T_3=\frac{2πr_3}{v_3}\)
Where,
\(r_3\)=Radius of the electron in \(n_3=3\)
\(=\frac{(n_3)^2h^2∈_0}{πme^2}\)
\(∴T_3=\frac{2πr_3}{v_3}\)
\(=\frac{2π×(3)^2×(6.62×10^{-34})^2×8.85×10^{-12}}{7.27×10^5×π \times  × 9.1×10^{-31}×(1.6×10^{-19})^2}\)
\(=4.12×10^{-15}s\)
Hence,the orbital period in each of these levels is \(1.52×10^{-16}s,1.22×10^{-15}s\), and \(4.12×10^{-15}s\) respectively.