Question

If work function of a metal is 3.7 eV and 5 eV of radiation falls on the metal surface then the kinetic energy of ejected photo electrons will be
$1) \: 1.38 \times {10}^{ - 18} j$
$2) \: 1.3 \times {10}^{ - 20} j$
$3) \: 3.2 \times {10}^{ - 18} j$
$4) \: 2.08 \times {10}^{ - 19} j$

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Answer 1

Answer:

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Answer 2

Given info : If work function of a metal is 3.7 eV and 5 eV of radiation falls on the metal surface.

To find : the kinetic energy of rejected photo electrons will be ...

solution : from photoelectric effect,

K.E = E - Φ

where E is incident energy and Φ is work function.

here, Work function of the metal , Φ = 3.7 eV

and incident energy, E = 5 eV

so, K.E = 5 - 3.7 = 1.3 eV

we know, 1 eV = 1.6 × 10^-19 J

K.E = 1.3 × 1.6 × 10^-19 J = 2.08 × 10^-19 J

Therefore the kinetic energy of ejected photo electrons will be 2.08 × 10^-19 J.

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