The work function of cesium metal is $2.14 e v .$ When the light of frequency $6 * 10^{1} 4$ Hz is incident on the metal surface photoemission of electrons occurs.What is the maximum kinetic energy of the emitted electrons.

0.35 e V .

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0.34 e V

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0.29 e V

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0.13 e V

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Solution

The correct option is A $0.35 e V .$
Given,
$ϕ = 2.14 e V$
$f = 6 \times 10^{14} H z$
The energy of photon,
$E = h f = \frac{6.6 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} e V$
$E = 2.485 e V$
From Einstein photoelectric equation,
$E = K_{m a x} + ϕ$
$K_{m a x} = E - ϕ = 2.485 e V - 2.14 e V$
$K_{m a x} = 0.35 e V$
The correct option is A.

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The work function of cesium metal is 2.14ev. When the light of frequency 6∗1014Hz is incident on the metal surface photoemission of electrons occurs.What is the maximum kinetic energy of the emitted electrons.

The correct option is A 0.35eV.Given,ϕ=2.14eVf=6×1014HzThe energy of photon, E=hf=6.6×10−34×6×10141.6×10−19eVE=2.485eVFrom Einstein photoelectric equation,E=Kmax+ϕKmax=E−ϕ=2.485eV−2.14eVKmax=0.35eVThe correct option is A.

The work function of cesium metal is $2.14 e v .$ When the light of frequency $6 * 10^{1} 4$ Hz is incident on the metal surface photoemission of electrons occurs.What is the maximum kinetic energy of the emitted electrons.