Light of wavelength 5000∘A is falling on a photo sensitive surface. If the surface has received 10−7J of energy, then the number of photons falling on the surface will be
A
5×1011
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B
2.5×1011
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C
9×1011
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D
1×1011
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Solution
The correct option is B2.5×1011 Energy received E=nhcλ⇒E=n(12400λ)eV⇒10−7=n×124005000×1.6×10−19⇒n=512.4×1.6×1012⇒n=2.5×1011