RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

    A)  \[0.55\overset{\text{o}}{\mathop{\text{A}}}\,\]          

    B)                         \[0.055\overset{\text{o}}{\mathop{\text{A}}}\,\]          

    C)         \[0.077\overset{\text{o}}{\mathop{\text{A}}}\,\]        

    D)                         \[0.095\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: B

    Solution :

    de Broglie wavelength \[\lambda =\frac{12.27}{\sqrt{V}}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[\lambda =\frac{12.27}{\sqrt{50000}}=0.055\overset{\text{o}}{\mathop{\text{A}}}\,\]


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