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Chemistry Question on Structure of atom

Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength 550 nm550\, nm is ..........., if 100 V100\, V and 1 A1\, A is supplied for one hour.

A

1×10241 \times 10^{24}

B

5×10245 \times 10^{24}

C

2×10242 \times 10^{24}

D

5×10235 \times 10^{23}

Answer

1×10241 \times 10^{24}

Explanation

Solution

E=hcλE =\frac{h c}{\lambda} E=(6.626×10−34Js)(3.0×108)550×10−9E =\frac{\left(6.626 \times 10^{-34} J s \right)\left(3.0 \times 10^{8}\right)}{550 \times 10^{-9}} E=0.036×10−17JE =0.036 \times 10^{-17} J Each proton carries 0.036×10−170.036 \times 10^{-17} J of energy. P=IV=1A×100 VP= IV =1 A \times 100\,V P=100 WP=100 \,W A 100 W100\, W source means that it outputs 100 J100\,J of energy every one hour. So, in one second, 100×3600 J100 \times 3600\, J of energy is liberated. Number of photons emitted == total energy output in one second/energy carried per photon n=100×36000.036×10−17=1×1024n=\frac{100 \times 3600}{0.036 \times 10^{-17}}=1 \times 10^{24}