A photocell in the saturation mode is irradiated by light of wavelength - -6600A- The corresponding spectral sensitivity of the cell is s -4-8 mA - W- Find the yield of photoelectrons- i-e- the number of photoelectrons produced by each incident photon- - Take-h-6-6- 10 -34 J-s -

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Standard XII

Physics

I vs V - Varying Frequency

A photocell i...

Question

A photocell in the saturation mode is irradiated by light of wavelength $λ = 6600 A$ . The corresponding spectral sensitivity of the cell is $s_{λ} = 4.8 m A / W$ . Find the yield of photoelectrons, i.e. the number of photoelectrons produced by each incident photon. $[T a k e : h = 6.6 \times 10^{- 34} J - s]$

9 \times 10^{- 2}

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9 \times 10^{- 4}

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9 \times 10^{- 3}

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Solution

The correct option is C $9 \times 10^{- 3}$
Let $n_{e}$ be the number of electrons, and $n_{p}$ be the number of photons.

Thus $s_{λ} = \frac{n_{e} e λ}{n_{p} h c}$

$⟹ \frac{n_{e}}{n_{p}} = \frac{s_{λ} h c}{e λ} = 9 \times 10^{- 3}$

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A photocell in the saturation mode is irradiated by light of wavelength λ=6600A. The corresponding spectral sensitivity of the cell is sλ=4.8mA/W. Find the yield of photoelectrons, i.e. the number of photoelectrons produced by each incident photon. [Take:h=6.6×10−34J−s]

The correct option is C 9×10−3Let ne be the number of electrons, and np be the number of photons.Thus sλ=neeλnphc⟹nenp=sλhceλ=9×10−3

The correct option is C $9 \times 10^{- 3}$
Let $n_{e}$ be the number of electrons, and $n_{p}$ be the number of photons.

Thus $s_{λ} = \frac{n_{e} e λ}{n_{p} h c}$

$⟹ \frac{n_{e}}{n_{p}} = \frac{s_{λ} h c}{e λ} = 9 \times 10^{- 3}$