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Q.

If the energy of a photon corresponding to a wavelength of 6000 Ao is 3.32×10−19J the photon energy for a wavelength of 4000 Ao will be

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a

1.4 eV

b

4.9 eV

c

3.1 eV

d

1.6 eV

answer is C.

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Detailed Solution

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

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