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Q.

The energy difference between two electronic states is 46.12 kcal/mol. What will be the frequency of the light emitted when an electron drops from the higher to the lower energy state ? (Planck's constant = 9.52  ×  10−14 kcal sec  mol−1)

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a

4.84  ×  1015  cycles  sec−1

b

4.84  ×  10−5  cycles  sec−1

c

4.84  ×  10−12  cycles  sec−1

d

4.84  ×  1014  cycles  sec−1

answer is D.

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Detailed Solution

ΔE=hcλE2−E1=hcλ⇒46.12KCal=9.52×10-14Kcal×3×108m/sλ1λ=ν-=46.12  ×  1033 ×  108 × 9.52 ×  10−14v=cλ  =3 ×  108 × 46.12  9.52  ×  10−14 × 3 × 108 ν=4.84  ×  1014  cycle/sec.
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