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Two identical photo cathodes receive light of frequencies $f_1\: and \: f_2$. If velocities of photoelectrons (of mass m) coming out are respectively $v_1\: and\: v_2$, then

$\begin {array} {1 1} (a)\;v_1^2-v_2^2=2h(f_1-f_2)/m & \quad (b)\;v_1+v_2= \bigg[ \large\frac{2h(f_1+f_2)}{m} \bigg]^{\large\frac{1}{2}} \\ (c)\;v_1^2+v_2^2=2h(f_1+f_2)/m & \quad (d)\;v_1-v_2= \bigg[ \large\frac{2h(f_1-f_2)}{m} \bigg]^{\large\frac{1}{2}} \end {array}$


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Ans : (a)
$ hf = hf_{\circ} + \large\frac{1}{2} mv^2$
Hence, $ v_1^2 = \large\frac{2hf_1}{m}- \large\frac{2hf_{\circ}}{m}$
$ v_2^2 = \large\frac{2hf_2}{m}- \large\frac{2hf_{\circ}}{m}$
So, $ v_1^2-v_2^2=2h \large\frac{(f_1-f_2)}{m}$


answered Dec 27, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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