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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Self Evaluation Test - Dual Nature of Radiation and Matter

  • question_answer
    When the minimum wavelength of X-rays is 2A then the applied potential difference between cathode and anticathode will be

    A) 6.2kV

    B) 2.48 kV

    C) 24.8kV

    D) 62kV

    Correct Answer: A

    Solution :

    [a] \[V=\frac{12400}{{{\lambda }_{\min }}\left( \overset{\text{o}}{\mathop{\text{A}}}\, \right)}Volt\Rightarrow V=\frac{12400}{2}=6.2KV\]


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