2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-2004

  • question_answer
    Light rays of wavelengths \[6000\,\overset{\text{o}}{\mathop{\text{A}}}\,\] and photon intensity \[39.6\text{ }watt/{{m}^{2}}\]is incident on a metal surface. If only one percent of photons incident on the surface emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:                 (Planck constant \[=\text{ }6.64\text{ }x\text{ }{{10}^{-34}}J-s,\] Velocity of light\[=3\times {{10}^{8}}m/s\])

    A)                 \[12\times {{10}^{18}}\]                               

    B)                  \[10\times {{10}^{18}}\]

    C)                 \[12\times {{10}^{17}}\]                               

    D)                  \[12\times {{10}^{16}}\]

    Correct Answer: C

    Solution :

                     Energy of each photon is                 \[E=\frac{hc}{\lambda }\]                 \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6000\times {{10}^{-10}}}\]                 \[=3.3\times {{10}^{-19}}J\]                 Now, power of source                 \[P=IA=39.6\times 1=39.6\,\text{watt}\]                 Number of photons emitted per second                 \[=\frac{p}{E}=\frac{39.6}{3.3\times {{10}^{-9}}}\]                 \[=12\times {{10}^{19}}\]                 Only 1% of photon eject elements, so no. of photoelectrons                 \[=12\times {{10}^{19}}\times \frac{1}{100}\]\[=12\times {{10}^{17}}\]


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