Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    Work function of a metal is 2.51 eV. Its threshold frequency is

    A)  \[5.9\times {{10}^{14}}cycle/s\]              

    B)  \[6.5\times {{10}^{14}}cycle/s\]

    C)  \[9.4\times {{10}^{14}}cycle/s\]              

    D)  \[6.08\times {{10}^{14}}cycle/s\]

    Correct Answer: D

    Solution :

    Work function \[\phi =h{{v}_{0}}\] (\[{{v}_{0}}\] = threshold frequency)                 So,          \[{{v}_{0}}=\frac{\phi }{h}=\frac{2.51\times 1.6\times {{10}^{-19}}}{6.6\times {{10}^{-34}}}\]                                 \[=6.08\times {{10}^{14}}cycle/s.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner