Explanation Given data: Terminal voltage of the battery 0.5 V. Power of the light is 1 W. Wavelength of the light is 600 nm. Percentage of the photons that give energy is 50%. Formula used: Write the expression for power. P in = E t (1) Here, P in is the power incident on the cell, E is the energy of the photon, and t is the time. Write the expression for total energy. E = N E photon (2) Here, N is the number of photon, and E photon is the energy of each photon. Write the expression for energy of photon and wavelength. E photon = h c λ (3) Here, h is the plank’s constant with value 6.63 × 10 − 34 J ⋅ s , c is the speed of light with value 3.0 × 10 8 m s , and λ is the wavelength. Write the expression for current and charge. I = Q t (4) Here, I is the current, and Q is the charge. Write the expression for charge and number of electron. Q = N e (5) Here, N is the number of electrons, and e is the charge of one electron. Write the expression for output power. P out = I V (6) Here, P out is the output power, and V is the voltage. Write the expression for efficiency. ε = P out P in (7) Here, ε is the efficiency, and P in is the input power. Combining equations (1), (2), and (3). P in = E t = N E photon t = h c λ ( N t ) photons ( N t ) photons = P in h c λ (8) Number of photons per second is ( N t ) Since only 50% of the photons give energy, so the number of photons per second is, ( N t ) electrons = 1 2 ( N t ) photons (9) Combining equation (4) and (5), I = N e t I = e ( N t ) electrons (10) Substitute 1.0 W for P in , 6.0 × 10 − 7 m for λ , 6

College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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