Question

A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the energy level that the transition originated. Transitions in the Balmer series all terminate n=2.
Delta E= -2.178 x10-18J ( 1/n2Final - 1/n2Initial )

The number is 5.

What is the energy of this transition in units of kJ/mole? ( hint: the anser is NOT 4.58x10-22kJ/mole or -4.58x10-22kJ/mole)

Answer 1

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$($\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$)

    (1/434 × $10^{-9}$) = -2.178 × $10^{-18}$ ($\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$)

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$)$\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$)$\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$  × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$)

    = (1.9878 × $10^{-25}$) ÷ (434 × $10^{-9}$)

    = 4.58 × $10^{-19}$ J

    = 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02×$10^{23}$, then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02×$10^{23}$)

         = 7.61 × $10^{-46}$ KJ/mole