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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    What will be the uncertainty in position (correct at 0.001%) of a electron which is moving with\[3.0\times {{10}^{4}}\]cm/s. Velocity, (mass of electron\[=9.1\times {{10}^{-28}},\text{ }A=6.626\times {{10}^{-22}}\] erg/s) Use the uncertainty principle of\[h/4\pi \]

    A)  3.84 cm          

    B)  1.92 cm

    C)  7.68 cm         

    D)  5.76 cm

    Correct Answer: B

    Solution :

     According   to   Heisenberg's   uncertainty principle \[\Delta x.m\Delta V\ge \frac{h}{4\pi }\] \[\Delta V=3\times {{10}^{4}}\times \frac{0.001}{100}=0.3\] \[\Delta x=\frac{h}{m\times \Delta V\times 4\pi }\] \[=\frac{6.626\times {{10}^{-27}}}{9.1\times {{10}^{-28}}\times 0.3\times 4\times 3.14}\] \[=1.92\text{ }cm\]


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