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NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #23Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.Ans : Given:
Wavelength of red light, `` \lambda `` = 653.1 nm = 653.1`` \times ``10`` -9`` m
Kinetic energy of H2 molecules `` \left(K\right)`` is given by
`` K=\frac{3}{2}kT`` ...(1)
Here, k = 8.62 × 10-5 eV/K
T = Temperature of H2 molecules
Energy released `` \left(E\right)`` when atom goes from ground state to n = 3 is given by
`` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
For ground state, n1 = 1
Also, n2 = 3
`` \therefore E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\left(\frac{8}{9}\right)...\left(2\right)``
`` ``
`` ``
Kinetic energy of H2 molecules = Energy released when hydrogen atom goes from ground state to n = 3 state
`` \therefore \frac{3}{2}\times 8.62\times {10}^{-5}\times T=\frac{13.6\times 8}{9}``
`` \Rightarrow T=\frac{13.6\times 8\times 2}{9\times 3\times 8.62\times {10}^{-5}}``
`` =9.4\times {10}^{4}\,\mathrm{\,K\,}``
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