Question

4.). An excited electron falls from n=4 to n=2.

a.) Calculate the energy change in Joules associated with this transition

b.) what Is the wavelength in nm of the emmitted electromagnetic radiation? what color Is it?

c.) calculate the energy change in Kj mol -1.

Answer #1

An excited electron falls from n = 4 to n = 2.

n1 = 4 and n2 = 2

(a). Energy change associated with this transition is:

E = - 13.6 (1 / (n2)^{2} - 1 / (n1)^{2})

= - 13.6 ((1 / 2^{2}) - (1 /
4^{2}))

= - 13.6 (0.25 - 0.0625)

= - 13.6 * 0.1875

= - 2.55 eV

= - 2.55 * 1.6x10^{-19}

**E = 4.08x10 ^{-19} J**

(b). Let the wavelength of the emmitted electromagnetic radiation be .

We know that:

E = h*c /

4.08x10^{-19} = 6.626x10^{-34} *
3x10^{8} /

=
6.626x10^{-34} * 3x10^{8} /
4.08x10^{-19}

=
4.872x10^{-7} m

= 487.2x10^{-9} m = 487.2 nm

Hence,

**Wavelength, = 487.2
nm.**

(c). Energy, E = 4.08x10^{-19} J

This is the energy of 1 electron.

1 mole = 6.023x10^{23} electrons.

Energy per mole = 4.08x10^{-19} J *
6.023x10^{23} electrons

= 2.457x10^{5} J/mole

**Energy per mole = 245.7 kJ/mole**

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