Ultraviolet radiation of 6.2 eV falls on an aluminium foil surface. Work function is 4.2 eV. The kinetic energy of the fastest electron emitted approximately

Question

Ultraviolet radiation of 6.2 eV falls on an aluminium foil surface. Work function is 4.2 eV. The kinetic energy of the fastest electron emitted approximately (A) 3.2 × 10-21 J (B) 3.2 × 10-19 J (C) 3.2 × 10-15 J (D) 3.2 × 10-17 J

Tardigrade Admin · Accepted Answer

From Einstein's relation, the kinetic energy

(K E)

of emitted electron is given by

K E = h v - W_{0}

where,

W_{0}

is work function of metal,

v

the frequency,

h

the Planck's constant.
Given,

h v = 6.2 e V, W_{0} = 4.2 e V

∴ K E = 6.2 - 4.2 = 2 e V

Also,

1 e V = 1.6 \times 1 0^{- 19} J

∴ K E = 2 \times 1.6 \times 1 0^{- 19} J

= 3.2 \times 1 0^{- 19} J

Q. Ultraviolet radiation of $6.2 e V$ falls on an aluminium foil surface. Work function is $4.2 e V$ . The kinetic energy of the fastest electron emitted approximately

$3.2 \times 1 0^{- 21} J$

$3.2 \times 1 0^{- 19} J$

$3.2 \times 1 0^{- 15} J$

$3.2 \times 1 0^{- 17} J$

Q. Ultraviolet radiation of 6.2eV falls on an aluminium foil surface. Work function is 4.2eV. The kinetic energy of the fastest electron emitted approximately

3.2×10−21J

3.2×10−19J

3.2×10−15J

3.2×10−17J

From Einstein's relation, the kinetic energy (KE) of emitted electron is given by KE=hv−W0​ where, W0​ is work function of metal, v the frequency, h the Planck's constant. Given, hv=6.2eV,W0​=4.2eV ∴KE=6.2−4.2=2eV Also, 1eV=1.6×10−19J ∴KE=2×1.6×10−19J =3.2×10−19J

Q. Ultraviolet radiation of $6.2 e V$ falls on an aluminium foil surface. Work function is $4.2 e V$ . The kinetic energy of the fastest electron emitted approximately

$3.2 \times 1 0^{- 21} J$

$3.2 \times 1 0^{- 19} J$

$3.2 \times 1 0^{- 15} J$

$3.2 \times 1 0^{- 17} J$