KVPY Sample Paper KVPY Stream-SX Model Paper-10

  • question_answer
    Let \[{{\lambda }_{1}}\] is the wavelength of an emitted photon in deexcitation of a Bohr's atom without considering recoil of atom. Also, \[{{\lambda }_{2}}\] is the wavelength of an emitted photon in deexcitation of a Bohr's atom considering recoil of atom. Then, correct statement is

    A) \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}>1\]                    

    B) \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=1\]

    C) \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}>\sqrt{2}\]                       

    D) \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}<1\]

    Correct Answer: D

    Solution :

    When an excited atom at rest in lab frame emits a photon of energy hf. the photon also carries a momentum \[p=\frac{hf}{c}.\]Conservation of momentum requires that atom also recoils with same momentum. So, energy carried by atom is \[\frac{{{p}^{2}}}{2m}.\]Hence, in emission we can write,\[\Delta E=hf+\] recoil energy of atom. So, emission energy of photon is smaller in case of recoil of atom. As \[E\propto \frac{1}{\lambda },\]emitted wavelength is larger. So, \[{{\lambda }_{2}}>{{\lambda }_{1}}\]and \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}<1.\]


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