A photon of energy 8-6 eV is incident on a metal surface of threshold frequency 1-6- 1015Hz- The maximum kinetic energy of the photoelectrons emitted in eV nearly

The correct option is C 2 K.E._max= incident energy work function =8.6 eV hν =(8.6 6.63× 10^ 34× 1.6× 10^151.6× 10^ 19 )eV =8.6 6.63 ...

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Standard XII

Physics

I vs V - Varying Frequency

A photon of e...

Question

A photon of energy 8.6 eV is incident on a metal surface of threshold frequency $1.6 \times 10^{15}$ Hz. The maximum kinetic energy of the photoelectrons emitted (in eV) nearly

1.6

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1.2

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Solution

The correct option is C 2
$K . E ._{m a x} =$ incident energy - work function
$= 8.6 e V - h ν$
$= (8.6 - \frac{6.63 \times 10^{- 34} \times 1.6 \times 10^{15}}{1.6 \times 10^{- 19}}$ )eV
$= 8.6 - 6.63$
$≃ 2 e V .$
So, the answer is option (C).

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A photon of energy 8.6 eV is incident on a metal surface of threshold frequency 1.6×1015Hz. The maximum kinetic energy of the photoelectrons emitted (in eV) nearly

The correct option is C 2K.E.max=incident energy - work function=8.6eV−hν=(8.6−6.63×10−34×1.6×10151.6×10−19)eV=8.6−6.63≃2eV.So, the answer is option (C).

A photon of energy 8.6 eV is incident on a metal surface of threshold frequency $1.6 \times 10^{15}$ Hz. The maximum kinetic energy of the photoelectrons emitted (in eV) nearly

The correct option is C 2
$K . E ._{m a x} =$ incident energy - work function
$= 8.6 e V - h ν$
$= (8.6 - \frac{6.63 \times 10^{- 34} \times 1.6 \times 10^{15}}{1.6 \times 10^{- 19}}$ )eV
$= 8.6 - 6.63$
$≃ 2 e V .$
So, the answer is option (C).