The threshold frequency v0 for a metal is 8 - 1014 s-1- What is the kinetic energy of an electron emitted having frequency v - 1-0 - 1015 s-1-

K.E = hv hϕ #160; = h(10× 10^14 8× 10^14) #160; = 6.626× 10^ 34(2×10^14) = 13.25× 10^ 20 J ...

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Standard XII

Chemistry

Photoelectric Effect

The threshold...

Question

The threshold frequency $v_{0}$ for a metal is $8 \times 10^{14} s^{- 1}$ . What is the kinetic energy of an electron emitted having frequency $v = 1.0 \times 10^{15} s^{- 1}$ .

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Solution

$K . E = h v - h ϕ$
$= h (10 \times 10^{14} - 8 \times 10^{14})$
$= 6.626 \times 10^{- 34} (2 \times 10^{14})$
$= 13.25 \times 10^{- 20} J$

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The threshold frequency v0 for a metal is 8×1014s−1. What is the kinetic energy of an electron emitted having frequency v=1.0×1015s−1.

K.E=hv−hϕ =h(10×1014−8×1014)=6.626×10−34(2×1014)=13.25×10−20J

The threshold frequency $v_{0}$ for a metal is $8 \times 10^{14} s^{- 1}$ . What is the kinetic energy of an electron emitted having frequency $v = 1.0 \times 10^{15} s^{- 1}$ .

$K . E = h v - h ϕ$
$= h (10 \times 10^{14} - 8 \times 10^{14})$
$= 6.626 \times 10^{- 34} (2 \times 10^{14})$
$= 13.25 \times 10^{- 20} J$