The photoelectric work function of a metal surface is 2eV- When light of frequency 1-5 -1015 Hz is incident on it- maximum kinetic energy of the photo-electrons- approximately is -

The correct option is D 4 eV K.E._max=hν ϕ = 6.6× 10^ 34× 1.5× 10^151.6× 10^ 19eV 2eV =6eV 2eV =4eV. So, the answer is option (D). ...

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Standard XII

Chemistry

Planck's Quantum Hypothesis

The photoelec...

Question

The photoelectric work function of a metal surface is 2eV. When light of frequency $1.5 \times 10^{15}$ Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

8 eV

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6 eV

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2 eV

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4 eV

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Solution

The correct option is D 4 eV
$K . E ._{m a x} = h ν - ϕ$

= $\frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} e V - 2 e V$

$= 6 e V - 2 e V$

$= 4 e V .$

So, the answer is option (D).

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The photoelectric work function of a metal surface is 2eV. When light of frequency 1.5×1015 Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

The correct option is D 4 eVK.E.max=hν−ϕ=6.6×10−34×1.5×10151.6×10−19eV−2eV=6eV−2eV=4eV.So, the answer is option (D).

The photoelectric work function of a metal surface is 2eV. When light of frequency $1.5 \times 10^{15}$ Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

The correct option is D 4 eV
$K . E ._{m a x} = h ν - ϕ$

= $\frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} e V - 2 e V$

$= 6 e V - 2 e V$

$= 4 e V .$

So, the answer is option (D).