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

Options

A. `4.84 xx 10^15 "cycles" sec^-1`
B. `4.84 xx 10^-5 "cycles" sec^-1`
C. `4.84 xx 10^-12 "cycles" sec^-1`
D. `4.84 xx 10^14 "cycles" sec^-1`

Answer

D

Explanation

(d) `Delta E = (hc)/(lamda)`
`Delta E_2 - Delta E_1 = (hc)/(lamda)`
`(46.12 xx 10^3)/(3 xx 10^8 xx 6.6 xx 10^-34) = n`
`v = (c)/(lamda) = (3 xx 10^8 xx 46.12 xx 10^3)/(6.6 xx 10^-34 xx 3 xx 10^8 xx N_A)`
`n = 4.84 xx 10^14 "cycle"//"sec"`.