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NCERT Class XII Chapter
Atoms
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Question : 7 of 18
Marks: +1, -0
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.
Solution:  
(a) Speed of the electron in Bohr’s nth orbit is
v = e22nhε0\frac{e^2}{2 n h \varepsilon_0} = α cn\frac{c}{n}
Speed of the electron in Bohr’s first (n = 1) orbit is
v1v_1 = 1137×3×1081\frac{1}{137} \times \frac{3 \times 10^8}{1} = 2.186 × 10ms−110 \text{ms}^{-1} = 2.186 × 106ms−110^6 \text{ms}^{-1}
v2v_2 = v12\frac{v_1}{2} = 1.093 × 106ms−110^6 \text{ms}^{-1}
v3v_3 = v13\frac{v_1}{3} = 0.729 × 106ms−110^6 \text{ms}^{-1}
(b) Orbital period of electron in Bohr’s first orbit is
T1T_1 = 2πr1v1\frac{2\pi r_1}{v_1} = 2×3.14×0.53×10−102.186×10\frac{2 \times 3.14 \times 0.53 \times 10^{-10}}{2.186 \times 10^{}} s = 1.52 × 10−1610^{-16} s
As TnT_n = n3T1n^3 T_1
∴ T2T_2 = (2)3(2)^3 × 1.52 × 10−1610^{-16} = 12.16 × 10−1610^{-16} = 1.22 × 10−1510^{-15} s
T3T_3 = (3)3(3)^3 × 1.52 × 10−1610^{-16} = 41.04 × 10−1610^{-16} = 4.10 × 10−1510^{-15} s.
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