Question

3. Complete it by calculating the AE's and wavelengths for all transitions indicated (non-shaded portions only) using the following equations = E 'n, high AE - Eow 'n,low 1.196x10 kJ nm / mol ΔΕ For example, the energy and the wavelength for the transition from Es to Es have been calculated and filled in the appropriate box. Units are omitted Elow n 2 n 3 n 4 n 5 n 6 n oo n 1 n 2 n 3 93 kJ 1282 nm n 4 n 5 n= 6

Answer 1

Answer -

Given,

$\Delta$E = E _high - E _low

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

Find Energy and wavelengths in nm.

According to Rydberg's Formula,

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

where, $\lambda$ = wavelength in m

R = Rydberg's constant (1.0974 * 10⁷ m^-1)

n = transition levels

So,

From n= 2 to 1

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/1²) - (1/2²)]

$\lambda$ = 121.5 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /121.5 nm

$\Delta$E = 984.36 kJ

Wavelength = 121.5 nm

Energy = 984.36 kJ

from 3 to 1

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/1²) - (1/3²)]

$\lambda$ = 102.5 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /102.5 nm

$\Delta$E = 1167.09 kJ

Wavelength = 102.5 nm

Energy = 1167.09 kJ

from 3 to 2

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/2²) - (1/3²)]

$\lambda$ = 656.11 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /656.11 nm

$\Delta$E = 182.33 kJ

Wavelength = 656.11 nm

Energy = 182.33 kJ

from 4 to 1

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/1²) - (1/4²)]

$\lambda$ = 97.20 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /97.20 nm

$\Delta$E = 1230.73 kJ

Wavelength = 97.20 nm

Energy = 1230.73 kJ

from 4 to 2

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/2²) - (1/4²)]

$\lambda$ = 486 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /486 nm

$\Delta$E = 246.14 kJ

Wavelength = 486 nm

Energy = 246.14 kJ

from 4 to 3

1/$\lambda$ = R $\cdot$[(1/n²_final) - (1/n²_initial)]

1/$\lambda$ = 1.0974 * 10⁷ m^-1$\cdot$[(1/3²) - (1/4²)]

$\lambda$ = 1874.61 nm

Also,

$\lambda$ = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\Delta$E

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) / $\lambda$

So,

$\Delta$E = (1.196 * 10⁵ kJ$\cdot$nm/mol) /1874.61 nm

$\Delta$E = 63.81 kJ

Wavelength = 1874.61 nm

Energy = 63.81 kJ

Similarly,

From n= 5 t n=1

Wavelength = 94.92 nm

Energy = 1260.29 kJ

From n= 5 t n=2

Wavelength = 433.94 nm

Energy = 275.67 kJ

From n= 5 t n=3

Wavelength = 1281.47 nm

Energy = 93.35 kJ

From n= 5 t n=4

Wavelength = 4050.07 nm

Energy = 29.54 kJ

From n= 6 t n=1

Wavelength = 93.73 nm

Energy = 1276.29 kJ

From n= 6 to n=2

Wavelength = 410.07 nm

Energy = 291.72 kJ

REST VALUES ARE CALCULATED AND FILLED IN THE TABLE.

	n = 2	n = 3	n = 4	n = 5	n = 6	n = $\infty$
n = 1	984.56kJ 121.5nm	1166.86kJ 102.52nm	1230.73 kJ 97.20 nm	1260.29 kJ 94.92 nm	1276.29 kJ 93.73 nm	1312.85kJ 91.12nm
n = 2	######	182.33 kJ 656.11nm	246.15 kJ 486 nm	275.67 kJ 433.94 nm	291.72 kJ 410.07 nm	328.20kJ 364.49nm
n = 3	######	######	63.81 kJ 1874.61nm	93.35 kJ 1281.47nm	109.39 kJ 1093.52nm	145.86kJ 820.12nm
n = 4	######	######	######	29.54 kJ 4050.07nm	45.58 kJ 2624.45nm	82.05kJ 1457.99nm
n = 6	######	######	######	######	9.67 kJ 12365.19nm	36.47kJ 3280.48nm
n = 7	######	######	######	######	######	26.79kJ 4465.09nm