JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    The de Broglie wavelength \[(\lambda )\]associated with a photoelectron varies with the frequency \[(\upsilon )\]of the incident radiation as, [\[{{\upsilon }_{0}}\] is threshold frequency] [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[\lambda \propto \frac{1}{(\upsilon -{{\upsilon }_{0}})}\]                        

    B) \[\lambda \propto \frac{1}{{{(\upsilon -{{\upsilon }_{0}})}^{1/2}}}\]

    C) \[\lambda \propto \frac{1}{{{(\upsilon -{{\upsilon }_{0}})}^{3/2}}}\]      

    D)   \[\lambda \propto \frac{1}{{{(\upsilon -{{\upsilon }_{0}})}^{1/4}}}\]

    Correct Answer: B

    Solution :

    For electron, \[\lambda =\frac{h}{\sqrt{2mK.E.}}\] \[kv=h{{v}_{0}}+K.E.\] \[K.E.=hv-h{{v}_{0}}\] \[\lambda =\frac{h}{\sqrt{2m(hv-h{{v}_{0}})}}=\frac{h}{\sqrt{2mh(v-{{v}_{0}})}}\] \[\lambda \propto \frac{1}{\sqrt{(v-{{v}_{0}})}}\]


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