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If wavelength of radiation is 4000Å=400n...

If wavelength of radiation is `4000Å=400nm` then the energy of the photon is

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The correct Answer is:
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`E=(hC)/(lambda)=(12400eVA^(0))/(4000A^(0))=(1240nm)/(400m)=3.1eV`
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NARAYNA-DUAL NATURE OF RADIATION AND MATTER-ASSERTION AND REASON
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