Which option correctly identifies metal that will produce the most energetic electrons and their energies ?

Lithium, 0.45 eV

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Tungsten, 1.75 eV

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Lithium, 2.30 eV

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Tungsten, 2.25 eV

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Solution

The correct option is A Lithium, 0.45 eV
Photoelectric effect:
$K E_{m a x} = h v - W_{o}$
$K E_{m a x} = \frac{h c}{λ} - W_{o}$
For most energetic atoms,
$W_{o}$ should me minimum
for $L i$ , $W_{o} = 2.3 e V$
$E = \frac{h c}{λ} = 2.75 e V$
$K E = 2.75 - 2.3 = 0.45 e V$

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Which option correctly identifies metal that will produce the most energetic electrons and their energies ?

The correct option is A Lithium, 0.45 eVPhotoelectric effect:KEmax=hv−WoKEmax=hcλ−WoFor most energetic atoms,Wo should me minimumfor Li, Wo=2.3eVE=hcλ=2.75eVKE=2.75−2.3=0.45eV

The correct option is A Lithium, 0.45 eV
Photoelectric effect:
$K E_{m a x} = h v - W_{o}$
$K E_{m a x} = \frac{h c}{λ} - W_{o}$
For most energetic atoms,
$W_{o}$ should me minimum
for $L i$ , $W_{o} = 2.3 e V$
$E = \frac{h c}{λ} = 2.75 e V$
$K E = 2.75 - 2.3 = 0.45 e V$