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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    A particle of mass M at rest decays into two masses \[{{m}_{1}}\] and \[{{m}_{2}}\] with non zero velocities. The ratio of de-Broglie wavelengths of the particles \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] is:

    A)  \[\frac{{{m}_{2}}}{{{m}_{1}}}\]                                

    B)  \[\frac{{{m}_{1}}}{{{m}_{2}}}\]

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

    D)  \[1:1\]

    Correct Answer: D

    Solution :

    By law of conservation of linear momentum                 \[{{m}_{1}}{{\upsilon }_{1}}={{m}_{2}}{{\upsilon }_{2}}\] So,          \[{{m}_{1}}{{\upsilon }_{1}}={{m}_{2}}{{\upsilon }_{2}}\] Now, de-Broglie wavelength \[\lambda =\frac{h}{m\upsilon }\] \[\therefore \]  \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{m}_{2}}{{\upsilon }_{2}}}{{{m}_{1}}{{\upsilon }_{1}}}\]                 \[{{\lambda }_{1}}:{{\lambda }_{2}}=1:1\]


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