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Questions  

If the energy of a photon corresponding to a wavelength of 6000 Ao is 3.32×1019J the photon energy for a wavelength of 4000 Ao will be

a
1.4 eV
b
4.9 eV
c
3.1 eV
d
1.6 eV

detailed solution

Correct option is C

E=hcλ⇒E1E2=λ2λ1⇒3.32×10−19E2=40006000E2=4.98×10−19J=3.1  eV

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