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Haryana PMT Haryana PMT Solved Paper-2009

  • question_answer
    42.          In a photoelectric experiment the relation between applied potential difference between cathode and anode V and the photoelectric                current I was found to be shown in graph below. If Plancks constant \[h=6.6\times {{10}^{-34}}J-s,\]the frequency of incident radiation would be nearly (in \[{{s}^{-1}}\])

    A)  \[0.436\times {{10}^{18}}\]                        

    B)  \[0.436\times {{10}^{17}}\]

    C)  \[0.775\times {{10}^{15}}\]                        

    D)  \[0.775\times {{10}^{16}}\]

    Correct Answer: C

    Solution :

                    For photoelectric effect, \[e{{V}_{0}}=hv\]                 \[\Rightarrow \]               \[v=\frac{e{{V}_{0}}}{h}\]                 \[\therefore \]  \[v=\frac{1.6\times {{10}^{-19}}\times 3.2}{6.6\times {{10}^{-34}}}\]                                 \[=0.775\times {{10}^{15}}Hz\]


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