Answer to Question #160132 in Physical Chemistry for KAARTIKA A/P ELONGKOVAN

Question

An electronic transition occurs between two neighboring energy levels and emits radiation with a wavelength of 432 nm. Determine n for the transition. Assume L = 1.2 nm

Answer 1

Energy expression for particle in a box

E_{n1 = (n1}² h²) \ 8mL²

En2 = (n2²h²) \ 8mL2

So now emission occoures so n1< n2

Electron jumps from n2 to n1

Energy difference = h²\8mL² (n2² - n1²)

m = 9.11 × 10^-31 kg

L = 1.2 nm = 1.2 × 10^-9 m

h = 6.626 × 10 ³⁴ J. S

Energy difference is = hv = hc\$\lambda$

$\lambda$ = 432 nm = 432 × 10^-9 m

C = 3 × 10⁸ m s-1

And n2 - n1 = 1

Due to very complex calculation the image is uploaded you can follow very easily and carefully