2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    Ultraviolet light wavelength 300 nm and intensity\[1.0\text{ }W{{n}^{-2}}\]falls on the surface of a photoelectric material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of\[1.0\text{ }c{{m}^{2}}\]of the surface is nearly

    A)  \[9.61\times {{10}^{14}}\]          

    B)  \[4.12\times {{10}^{13}}\]

    C)  \[1.51\times {{10}^{12}}\]          

    D)  \[2.13\times {{10}^{11}}\]

    Correct Answer: C

    Solution :

                     Energy of each photon,                 \[E=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{300\times {{10}^{-9}}}\]                 \[E=6.6\times {{10}^{-19}}J\]                 Power of source                 \[P=IA=1.0\times 1.0\times {{10}^{-4}}={{10}^{-4}}W\]                 So, number of photons per sec as                 \[\frac{N}{t}=\frac{P}{E}=\frac{{{10}^{-4}}}{6.6\times {{10}^{-19}}}\]                 \[\Rightarrow \]               \[N'=1.51\times {{10}^{12}}/s\]                 Number of electrons emitted is                 \[N'=\frac{1}{100}\times \frac{{{10}^{-4}}}{6.6\times {{10}^{-19}}}\]        \[\Rightarrow \]                \[N'=1.51\times {{10}^{12}}/s\]


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