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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If\[{{\lambda }_{\min }}\] is the smallest possible wavelength of X-ray in the spectrum, the variation of \[\log {{\lambda }_{\min }}\] with log V is correctly represented in:                             JEE Main Solved Paper-2017

    A)        

    B)  

    C)          

    D)  

    Correct Answer: C

    Solution :

     \[\frac{hc}{{{\lambda }_{\min }}}=eV\]                 \[\frac{1}{{{\lambda }_{\min }}}=\frac{eV}{hc}\] \[\ell n\left( \frac{1}{{{\lambda }_{\min }}} \right)=\ell nV+\ell n\frac{e}{hc}\] \[-n({{\lambda }_{\min }})=\ell nV+\ell n\frac{e}{hc}\] \[\ell n({{\lambda }_{\min }})=-\ell nV-\ell n\left( \frac{e}{hc} \right)\] It is a straight line with -ve slope.


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