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JEE Advanced 2018 Paper 2

Section: Physics

Question : 13 of 54

Marks: +1, -0

6.2\,\text{eV}

F=n\times10^{-4}\,\text{N}

m_e=9\times10^{-31}\,\text{kg}

1.0\,\text{eV}=1.6\times10^{-19}\,\text{J}

Solution:

power = nhv n= number of protons per second

Since KE=0

hc=\phi

200=n[6.25\times16\times10^{-19}\,\text{joule}]\;\;\;\;\;\;n=\frac{200}{1.6\times10^{-19}\times6.25}

As photon just above threshold frequency

KE_{\text{max}}

is zero and they are accelerated by potential difference of 500V.

KE_f=q\Delta V

\frac{P^2}{2m}=q\Delta V\Rightarrow P=\sqrt{2m\Delta V}

Since efficiency is 100%, number of electrons = number of photons per second As photon is completely absorbed force exerted = nmv

=\frac{200}{6.25\times1.6\times10^{-19}}\times\sqrt{2(9\times10^{-31})\times1.6\times10^{-19}\times500}

=\frac{3\times200\times10^{-25}\times\sqrt{1600}}{6.25\times1.6\times10^{-19}}=\frac{2\times40}{6.25\times1.6}\times10^{-4}\times3=24

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