UV light radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of faster electrons emitted is approximately

3.2 \times 10^{- 15} J

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3.2 \times 10^{- 17} J

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3.2 \times 10^{- 19} J

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3.2 \times 10^{- 21} J

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Solution

The correct option is C $3.2 \times 10^{- 19} J$
KE of emitted electron is
$K_{m a x} = h v - W$
= 6.2 eV - 4.2 eV
= 2 eV
$= 2 \times 1.6 \times 10^{- 19}$
$= 3.2 \times 10^{- 19} J$

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Q. 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

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UV light radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of faster electrons emitted is approximately

The correct option is C 3.2×10−19JKE of emitted electron is Kmax=hv−W = 6.2 eV - 4.2 eV = 2 eV =2×1.6×10−19 =3.2×10−19J

The correct option is C $3.2 \times 10^{- 19} J$
KE of emitted electron is
$K_{m a x} = h v - W$
= 6.2 eV - 4.2 eV
= 2 eV
$= 2 \times 1.6 \times 10^{- 19}$
$= 3.2 \times 10^{- 19} J$