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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    Ultraviolet light of wavelengths \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy \[K.{{E}_{1}}\] and \[K.{{E}_{2}}\] respectively. Find the value of Planck's constant.   

    A)  \[h=\left| \frac{\left( K.{{E}_{2}}-K.{{E}_{1}} \right)\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)}{C\left( {{\lambda }_{1}}-{{\lambda }_{2}} \right)} \right|\]

    B)  \[h=\left| \frac{\left( K.{{E}_{1}}-K.{{E}_{2}} \right)\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)}{C{{\lambda }_{1}}{{\lambda }_{2}}} \right|\]

    C)  \[h=\left| \frac{\left( K.{{E}_{1}}-K.{{E}_{2}} \right){{\lambda }_{1}}{{\lambda }_{2}}}{C\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)} \right|\]

    D)        None of These

    Correct Answer: C

    Solution :

    [c] \[\frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{0}}}=K.E{{.}_{1}}\text{ and }\frac{hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}=K.E{{.}_{2}}\] \[\Rightarrow \frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{2}}}=K.E{{.}_{1}}-K.E{{.}_{2}}\] \[\Rightarrow hc\left[ \frac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}} \right]=K.E{{.}_{1}}-K.E{{.}_{2}}\] \[\therefore h=\frac{\left( K.E{{.}_{1}}-K.E{{.}_{2}} \right){{\lambda }_{1}}{{\lambda }_{2}}}{c\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)}\]


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