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

3.2 \times 10^{- 21}

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3.2 \times 10^{- 19}

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3.2 \times 10^{- 17}

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3.2 \times 10^{- 15}

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Solution

The correct option is B $3.2 \times 10^{- 19}$
$K_{m a x} (e V) = E (e V) - W_{0} (e V) = 6.2 - 4.2 = 2 e V$
$∴ K_{m a x} (J o u l e s) = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19}$

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

The correct option is B 3.2×10−19Kmax(eV)=E(eV)−W0(eV)=6.2−4.2=2 eV ∴ Kmax(Joules)=2×1.6×10−19J=3.2×10−19

The correct option is B $3.2 \times 10^{- 19}$
$K_{m a x} (e V) = E (e V) - W_{0} (e V) = 6.2 - 4.2 = 2 e V$
$∴ K_{m a x} (J o u l e s) = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19}$