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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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A photocell is operating in saturation mode with a aphotocurrent 4.8mA when a monochromatic radiation of wavelength $\;3000A^{0}\;$ and power 1mW is incident. When another monochromatic radiation of wavelength $\;1650A^{0}\;$ and power 5mW is incident , it is observed that maximum velocity of photoelectron generation per incident to be same for both cases . Than the threshold wavelength for cell is :

$(a)\;5268A^{0}\qquad(b)\;3128A^{0}\qquad(c)\;1264A^{0}\qquad(d)\;4133A^{0}$

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Answer : (d) $\;4133 A^{0}$
Explanation :
$K_{1}=\large\frac{12400}{3000}-W=4.13-W----->(1)$
$K_{2}=\large\frac{12400}{1650}-W=7.51-W----->(2)$
Since $\;V_{2}=2V_{1} \quad \; S0\;,K_{2}=4K_{1}----->(3)$
Solving above equations .
$W=3eV$
Therefore , Threshold wavelength $\;\lambda_{0}=\large\frac{12400}{3}=4133A^{0}\;.$
answered Feb 22, 2014 by yamini.v
 

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