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

Tardigrade Admin · Accepted Answer

Using the relation KE=hv-ω 0 KE=6.2-4.2 =2 eV =2× 1.6× 10-19=3.2× 10-19 J

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

$3.2 \times 10^{- 21} J$

$3.2 \times 10^{- 19} J$

$3.2 \times 10^{- 15} J$

$3.2 \times 10^{- 17} J$

Using the relation $K E = h v - ω_{0}$ $K E = 6.2 - 4.2 = 2 e V$ $= 2 \times 1.6 \times 10^{- 19} = 3.2 \times 10^{- 19} J$

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

3.2×10−21J

3.2×10−19J

3.2×10−15J

3.2×10−17J

Using the relation KE=hv−ω0​ KE=6.2−4.2=2eV =2×1.6×10−19=3.2×10−19J

$3.2 \times 10^{- 21} J$

$3.2 \times 10^{- 19} J$

$3.2 \times 10^{- 15} J$

$3.2 \times 10^{- 17} J$

Using the relation $K E = h v - ω_{0}$ $K E = 6.2 - 4.2 = 2 e V$ $= 2 \times 1.6 \times 10^{- 19} = 3.2 \times 10^{- 19} J$