$1.5 m W$ of 400\ nm light is directed at a photoelectric cell. If $0.10$ percent fo the incident photons produce photoelectrons, then the current in the cell is

4.8 μ A

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48 μ A

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1.8 μ A

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0.48 μ A

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Solution

The correct option is D $0.48 μ A$
For wavelength $400 n m$ , energy of each photon be $E = \frac{h_{c}}{λ}$ , $n$ number of photon will energy
$n (E) = 1.5 m w$
$0.1$ % photons extract electrons.
Now, $n q = n e$
$n q = 1.5 \times \frac{1}{100} = 0.48 μ A$ .

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1.5 mW of 400\ nm light is directed at a photoelectric cell. If 0.10 percent fo the incident photons produce photoelectrons, then the current in the cell is

The correct option is D 0.48μAFor wavelength 400nm, energy of each photon be E=hcλ, n number of photon will energy n(E)=1.5mw0.1% photons extract electrons.Now, nq=nenq=1.5×1100=0.48μA.

$1.5 m W$ of 400\ nm light is directed at a photoelectric cell. If $0.10$ percent fo the incident photons produce photoelectrons, then the current in the cell is

The correct option is D $0.48 μ A$
For wavelength $400 n m$ , energy of each photon be $E = \frac{h_{c}}{λ}$ , $n$ number of photon will energy
$n (E) = 1.5 m w$
$0.1$ % photons extract electrons.
Now, $n q = n e$
$n q = 1.5 \times \frac{1}{100} = 0.48 μ A$ .