Light of wavelength 5000 , mathring{A} falls | vedclass

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(B) The energy of a single photon is given by $E_p = \frac{hc}{\lambda}$.Given: $h = 6.63 	imes 10^{-34} \, 	ext{J s}$,$c = 3 	imes 10^8 \, 	ext{m

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$2.5 	imes 10^{11}$

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(B) The energy of a single photon is given by $E_p = \frac{hc}{\lambda}$.Given: $h = 6.63 	imes 10^{-34} \, 	ext{J s}$,$c = 3 	imes 10^8 \, 	ext{m/s}$,and $\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, 	ext{m} = 5 	imes 10^{-7} \, 	ext{m}$.$E_p = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{5 	imes 10^{-7}} \approx 3.978 	imes 10^{-19} \, 	ext{J} \approx 4 	imes 10^{-19} \, 	ext{J}$.Let $n$ be the number of photons. The total energy $E_{total} = n 	imes E_p$.Given $E_{total} = 10^{-7} \, 	ext{J}$.$n = \frac{E_{total}}{E_p} = \frac{10^{-7}}{4 	imes 10^{-19}} = 0.25 	imes 10^{12} = 2.5 	imes 10^{11}$.Thus,the number of photons is $2.5 	imes 10^{11}$.

Light of wavelength $5000 \, \mathring{A}$ falls on a sensitive surface. If the surface has received $10^{-7} \, \text{J}$ of energy,then the number of photons falling on the surface will be:

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