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Question

Chemistry Question on Structure of atom

A 100100 watt bulb emits monochromatic light of wavelength 400 nm400\,nm. Calculate the number of photons emitted per second by the bulb.

A

3×1020 s−13 \times 10^{20}\, s^{-1}

B

2×10−20 s−12 \times 10^{-20}\, s^{-1}

C

2×1020 s−12 \times 10^{20}\, s^{-1}

D

1×10−20 s−11 \times 10^{-20}\, s^{-1}

Answer

2×1020 s−12 \times 10^{20}\, s^{-1}

Explanation

Solution

Power of the bulb =100= 100 watt =100 J s−1= 100 \,J\,s^{-1} Energy of one photon E=hυ=hc/λE = h\upsilon = hc/\lambda =6.626×10−34 J s×3×108 m s−1400×10−9 m=\frac{6.626 \times 10^{-34}\,J\,s \times 3 \times 10^{8}\,m\,s^{-1}}{400 \times 10^{-9}\,m} =4.969×10−19 J=4.969\times10^{-19}\,J Number of photons emitted =100 J s−14.969×10−19 J=\frac{100\,J\,s^{-1}}{4.969 \times 10^{-19}\,J} =2.012×1020 s−1=2.012\times10^{20}\,s^{-1}