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The energy of a photon is given as Delta...

The energy of a photon is given as `Delta E//` atom `3.03xx10^(-19) J "atom"^(-1)`. Then the wavelength `(lamda)` of the photon is

A

65.6nm

B

65.6nm

C

0.656 nm

D

6.56 nm

Text Solution

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The correct Answer is:
B
`Delta E= hv=(hc)/(lamda)`,
`therefore lamda=(hc)/(Delta E)=(6.63xx10^(-34)(3xx10^(8)))/(3.03xx10^(-19))=656 nm`
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