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A 200 W bulb emits monochromatic light o...

A 200 W bulb emits monochromatic light of wavelenght 1400 A and only 10% of the energy is emitted as light. The number of photons emitted by the bulb per second will be

A

`1.4` x `10^(18)`

B

`1.4` x `10^(20)`

C

`1.4` x `10^(19)`

D

`1.4` x `10^(21)`

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The correct Answer is:
To solve the problem, we need to calculate the number of photons emitted by the bulb per second. Here’s a step-by-step breakdown of the solution: ### Step 1: Calculate the total power emitted as light The bulb has a power rating of 200 W, but only 10% of this power is emitted as light. \[ \text{Power emitted as light} = 200 \, \text{W} \times 0.10 = 20 \, \text{W} \] ### Step 2: Convert power to energy per second Since 1 W = 1 J/s, the energy emitted as light per second is: \[ \text{Energy emitted per second} = 20 \, \text{J/s} \] ### Step 3: Calculate the energy of a single photon The energy of a photon can be calculated using the formula: \[ E = \frac{hc}{\lambda} \] Where: - \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant) - \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light) - \( \lambda = 1400 \, \text{Å} = 1400 \times 10^{-10} \, \text{m} \) Now substituting the values: \[ E = \frac{(6.63 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{1400 \times 10^{-10} \, \text{m}} \] Calculating the numerator: \[ E = \frac{(6.63 \times 3) \times 10^{-34 + 8}}{1400 \times 10^{-10}} = \frac{19.89 \times 10^{-26}}{1400 \times 10^{-10}} = \frac{19.89}{1400} \times 10^{-16} \] Calculating further: \[ E \approx 1.42 \times 10^{-19} \, \text{J} \] ### Step 4: Calculate the number of photons emitted per second Now we can find the number of photons emitted per second using the formula: \[ \text{Number of photons} = \frac{\text{Total energy emitted per second}}{\text{Energy of one photon}} \] Substituting the values: \[ \text{Number of photons} = \frac{20 \, \text{J}}{1.42 \times 10^{-19} \, \text{J}} \approx 1.41 \times 10^{20} \] ### Final Answer The number of photons emitted by the bulb per second is approximately \( 1.41 \times 10^{20} \). ---
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