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Ultraviolet radiation of 6.2 eV falls on an aluminum surface (work function 4.2 eV). The kinetic energy in joule of the fastest electron emitted is approximately

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$3 \times 10^{- 21}$

$3 \times 10^{- 19}$

$3 \times 10^{- 17}$

$3 \times 10^{- 12}$

answer is B.

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Detailed Solution

$h v = W_{0} + K . E 6 \cdot 2 = 4 \cdot 2 + K \cdot E . or K \cdot E . = 2 e V o r K . E . = 2 \times (1 \cdot 6 \times 10^{- 34}) J = 3 \cdot 2 \times 10^{- 19} J$

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