Question

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(a) 3.2×10⁻²¹(b) 3.2×10⁻¹⁹
(c) 3.2×10⁻¹⁷(d) 3.2×10⁻¹⁵

Answer 1

we know that Total energy is equal to work function + kinetic energy

so we can say that

E = W +K.E

so , computing this in the given values

we shall get ,

6.4 = 4.2 + K.E

K.E = 3.2 eV

converting it to joules

1eV is equal to 1 ×10^-19 joules
( rounded off )


so K.E = 3.2 into 1 x 10 to power minus 19 joules

Then after computing the total value is

3.2 ×10^-19

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Answer 2

Answer:3.2×10^-19J

Explanation:

E-w.=6.2-4.2=2.0ev

2×1.6×10^-19

=3.2×10^-19J