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Physics Question on Atoms

(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

(a) Let v1v_1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1=1n_1=1. For charge (e) of an electron, v1v_1 is given by the relation,
v1=e2n14π∈0(h2π)=e22∈0hv_1=\frac{e^2}{n_14π∈_0(\frac{h}{2π})}=\frac{e^2}{2∈_0h}
Where, e=1.6×10−19Ce=1.6×10^{-19}C
∈0∈_0=Permittivity of free space=8.85×10−12N−1C2m−2=8.85×10^{-12}N^{-1}C^2m^{-2}
h=planck's constant=6.62×10−34Js=6.62×10^{-34}Js
∴v1=(1.6×10−19)22×8.85×10−12×6.62×10−34∴v_1=\frac{(1.6×10^{-19})^2}{2×8.85×10^{-12}×6.62×10^{-34}}
=0.0218×108=2.18×106m/s=0.0218×10^8=2.18×10^6m/s
For level n2=2n_2=2, we can write the relation for thecorresponding orbital speed as:
v2=e2n22∈0hv_2=\frac{e^2}{n^2_2∈_0h}
=(1.6×10−19)23×2×8.85×10−12×6.62×10−34=\frac{(1.6×10^{-19})^2}{3×2×8.85×10^{-12}×6.62×10^{-34}}
=1.09×106m/s=1.09×10^6m/s
And, for n3=3n_3=3, we can write the relation for the corresponding orbital speed as:
v3=e2n32∈0hv_3=\frac{e^2}{n_32∈_0h}
=(1.6×10−19)23×2×8.85×10−12×6.62×10−34=\frac{(1.6×10^{-19})^2}{3×2×8.85×10^{-12}×6.62×10^{-34}}
=7.27×105m/s=7.27×10^5m/s
Hence,the speed of the electron in a hydrogen atom in n=1, n=2, and n=3 is 2.18×106m/s,1.09×106m/s,7.27×105m/s2.18×10^6m/s,1.09×10^6m/s,7.27×10^5m/s respectively.

(b)Let T1T_1 be the orbital peroid of the electron when it is in level n1=1n_1=1.
Orbital period is related to orbital speed as:
T1=2πr1v1T_1=\frac{2πr_1}{v_1}
Where, r1r_1=Radius of the orbit=n12h2∈0πme2=\frac{n_1^2h^2∈_0}{πme^2}
h=planck's constant=6.62×10−34Js=6.62×10^{-34}Js
e=charge on an electron=1.6×10−19C=1.6×10^{-19}C
∈0=∈_0=permittivity of free space=8.85×10−12N−1C2m−2=8.85×10^{-12}N^{-1}C^2m^{-2}
m=Mass of an electron=9.1×10−31kg=9.1×10^{-31}kg
∴T1=2πr1v1∴T_1=\frac{2πr_1}{v_1}
=2π×(1)2×(6.62×10−34)2×8.85×10−122.18×106×π×9.1×10−31×(1.6×10−19)2=\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=1.527×10−16s=15.27×10^{-17}=1.527×10^{-16}s
For level n2=2n_2=2, we can write the period as:
T2=2πr2v2T_2=\frac{2πr_2}{v_2}
Where r2r_2=Radius of the electron in n2=2n_2=2
=(n2)2h2∈0πme2=\frac{(n_2)^2h^2∈_0}{πme^2}
∴T2=2πr2v2∴T_2=\frac{2πr_2}{v_2}
=2π×(2)2×(6.62×10−34)2×8.85×10−121.09×106×π×9.1×10−31×(1.6×10−19)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}
=1.22×10−15s=1.22×10^{-15}s
And, for level n3=3n_3=3, we can write the period as:
T3=2πr3v3T_3=\frac{2πr_3}{v_3}
Where,
r3r_3=Radius of the electron in n3=3n_3=3
=(n3)2h2∈0πme2=\frac{(n_3)^2h^2∈_0}{πme^2}
∴T3=2πr3v3∴T_3=\frac{2πr_3}{v_3}
=2π×(3)2×(6.62×10−34)2×8.85×10−127.27×105×π××9.1×10−31×(1.6×10−19)2=\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}
=4.12×10−15s=4.12×10^{-15}s
Hence,the orbital period in each of these levels is 1.52×10−16s,1.22×10−15s1.52×10^{-16}s,1.22×10^{-15}s, and 4.12×10−15s4.12×10^{-15}s respectively.