Question

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an electron drops from higher to lower state. Planck's constant , h = 39.79 × 1/10¹⁴ kJ sec per mol .​

Answer 1

`{\qquad\qquad\huge\underline{{\sf Answer}}}`

Here we go ~

Energy difference btween the two electronic states can be expressed as :

`{ \qquad \sf  \dashrightarrow \: \Delta E = h\\u}`

[ h = planks constant,
`{\: \\u }`= frequency ]

`\qquad \sf  \dashrightarrow \:214.68 = 39.79 * 10 {}^( - 14) * \\u`

`\qquad \sf  \dashrightarrow \: \\u = \cfrac{214.68}{39.79 * 10 {}^( - 4) }`

`\qquad \sf  \dashrightarrow \: \\u = \cfrac{214.68}{39.79 } * 10 {}^(14)`

`\qquad \sf  \dashrightarrow \: \\u \approx 5.395 *10 {}^(14) \:\:hertz`