Light of wavelength 200 nm shines on an aluminium, $4.20 eV$ is required to eject an electron. What is the kinetic energy of the fastest ejected electrons in eV?
(take $h = 6.6 \times 10^{- 34} J s$ )

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Solution

Using the Einstein's equation of photoelectric effect,
$\frac{h c}{λ} = K . E . + ϕ$
$\Rightarrow K . E . = \frac{h c}{λ} - ϕ$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{200 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 4.2$

$K . E . = 1.9875 \approx 1.99 eV$

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Light of wavelength 200 nm shines on an aluminium, 4.20 eV is required to eject an electron. What is the kinetic energy of the fastest ejected electrons in eV? (take h=6.6×10−34 Js)

Using the Einstein's equation of photoelectric effect, hcλ=K.E.+ϕ ⇒K.E.=hcλ−ϕ =6.6×10−34×3×108200×10−9×1.6×10−19−4.2 K.E.=1.9875≈1.99 eV

Light of wavelength 200 nm shines on an aluminium, $4.20 eV$ is required to eject an electron. What is the kinetic energy of the fastest ejected electrons in eV?
(take $h = 6.6 \times 10^{- 34} J s$ )

Using the Einstein's equation of photoelectric effect,
$\frac{h c}{λ} = K . E . + ϕ$
$\Rightarrow K . E . = \frac{h c}{λ} - ϕ$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{200 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 4.2$

$K . E . = 1.9875 \approx 1.99 eV$