NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #14A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?digAnsr: bAns : Energy (E) of the nth state of hydrogen atom is given by
E = `` \frac{13.6}{{n}^{2}}`` eV
For n = 6,
`` \therefore E=\frac{-13.6}{36}=-0.377777777`` eV
Energy of hydrogen atom in the ground state = -13.6 eV
Energy emitted in the second transition = Energy of ground state `` -`` (Energy of hydrogen atom in the 6th state + Energy of photon)
`` =13.6-\left(0.3777\overline{)7}+1.13\right)``
`` =12.09=12.1\,\mathrm{\,eV\,}``
(b)
Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state)
= 1.13 eV + 0.377 eV
= 1.507 eV
Energy of the nth state can expressed as
`` \frac{13.6}{{n}^{2}}=1.507``
`` ``
`` \therefore n=\sqrt{\frac{13.6}{1.507}}=\sqrt{9.024}\approx 3``
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