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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 9 of 27
Marks: +1, -0
State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.
OR
Calculate the orbital period of the electron in the first excited state of hydrogen atom.
Solution:  
Statement of Bohr's quantization condition
Calculation of shortest wavelength
Identification of part of electromagnetic spectrum
Electron revolves around the nucleus only in those orbits for which the angular momentum is some integral of h2Ï€\frac{h}{2\pi} . (where hh is Planck's constant)
Also give full credit if a student write mathematically mvr=nh2Ï€m v r = \frac{n h}{2\pi}
1λ=R(1nf2−1ni2)\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
For Brackett Series,
Shortest wavelength is for the transition of electrons from ni=∞n_i = \infty to nf=4n_f = 4
1λ=R(142)=R16\frac{1}{\lambda}=R\left(\frac{1}{4^2}\right)=\frac{R}{16}
λ=16R m\lambda = \frac{16}{R}\,\text{m}
=1458.5 nm on substitution of value of R1=1458.5\,\text{nm}\,\text{on substitution of value of } R 1
OR
Statement of the formula for rnr_n
Statement of the formula for vnv_n
Obtaining formula for TnT_n
Getting expression for T2(n=2)T_2 (n=2)
Radius, rn=h2ϵ0πme2n2r_n = \frac{h^2 \epsilon_0}{\pi m e^2} n^2
velocity, vn=2πe24πϵ0h1nv_n = \frac{2\pi e^2}{4\pi \epsilon_0 h} \frac{1}{n}
Timeperiod, Tn=2πrnvn=4ϵ02h3n3me4T_n = \frac{2\pi r_n}{v_n} = \frac{4\epsilon_0^2 h^3 n^3}{m e^4}
For first excited state of hydrogen atom n=2n=2
T2=32ϵ02h3me4T_2 = \frac{32\epsilon_0^2 h^3}{m e^4}
On calculation we get T2≈1.22×10−15 sT_2 \approx 1.22 \times 10^{-15}\,\text{s}.
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