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The work function of a certain metal is 3.31 × 10^–19 J. Then the maximum kinetic energy of photo-electrons emitted by incident radiation of wave length 5000 $\overset{0}{A}$ is (given h = 6.62×10^–34 J -s, C=3×10⁸ms^–1, e = 1.6×10^–19 coul.)

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2. 48 eV

0. 41 eV

2. 07 eV

0. 82 eV

answer is B.

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Detailed Solution

$K{E_{\max }} = E - {W_0} = \frac{{hc}}{\lambda } - {W_o}\$

$= \frac{{12400eV{\mathop A\limits^0 }}}{{5000{\mathop A\limits^0 }}} - \frac{{3.31 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}eV\$

= 2.48eV – 2.07eV = 0.41eV

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