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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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In an experiment on photo electric emission , following observations were made ; (1) Wavelength of incident light = $\;1.98\times10^{-7}m\;$ 2 Stopping potential = $\;2.5 volt\;$. Then kinetic energy of photoelectrons with maximum speed and Thus hold frequency are respectively

$(a)\;5eV,9\times10^{14} Hz\qquad(b)\;2.5eV,9\times10^{14} Hz\qquad(c)\;5eV,18\times10^{14} Hz\qquad(d)\;2eV,18\times10^{14} Hz$

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Answer : (b) $\;2.5eV,9\times10^{14} Hz$
Explanation :
Since $\;V_{s}=2.5V , \quad \; K_{max}=eV_{s}$
So , $\;K_{max}=2.5 eV$
Energy of incident photon
$\varepsilon =\large\frac{12400}{1980}eV=6.26eV$
$W=E-K_{max}=3.76eV$
$h \nu_{th}=W=3.76\times1.6\times10^{-19}J$
$\nu_{th}=\large\frac{3.76\times1.6\times10^{-19}}{6.6\times10^{-34}} \approx 9.1\times10^{14} Hz\;.$
answered Feb 22, 2014 by yamini.v
 

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