Respuesta :
Explanation:
Using Reydberg's equation, we will calculate the energy emitted by the photon for the given transitions.
The equation is as follows.
[tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
1). For [tex]n_{1}[/tex] = 7, [tex]n_{2}[/tex] = 1, Z for H = 1 [tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (1)^{2}[\frac{1}{(1)^{2}} - \frac{1}{(7)^{2}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (1 - 0.02)[/tex]
= [tex]-12.13 \times 10^{-18} J[/tex]
As, E = [tex]\frac{hc}{\lambda}[/tex]
Putting the given values and calculate the wavelength as follows.
E = [tex]\frac{hc}{\lambda}[/tex]
[tex]12.13 \times 10^{-18} J[/tex] = [tex]\frac{6.626 \times 10^{-34}Js \times 3 \times 10^{8 m/s}}{\lambda}[/tex]
[tex]\lambda = 1.58 \times 10^{-8}[/tex] m
= [tex]15.8 \times 10^{-9}[/tex] m
= 15.8 nm (as 1 m = [tex]10^{-9}[/tex] nm)
2). For [tex]n_{1}[/tex] = 7, [tex]n_{2}[/tex] = 5, Z for H = 1 [tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (1)^{2}[\frac{1}{(5)^{2}} - \frac{1}{(7)^{2}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (0.04 - 0.02)[/tex]
= [tex]-0.0435 \times 10^{-18} J[/tex]
As, E = [tex]\frac{hc}{\lambda}[/tex]
Putting the given values and calculate the wavelength as follows.
E = [tex]\frac{hc}{\lambda}[/tex]
[tex]0.0435 \times 10^{-18} J[/tex] = [tex]\frac{6.626 \times 10^{-34}Js \times 3 \times 10^{8 m/s}}{\lambda}[/tex]
[tex]\lambda = 456.9 \times 10^{-8}[/tex] m
= 45.6 nm
3). For [tex]n_{1}[/tex] = 1, [tex]n_{2}[/tex] = 7, Z for H = 1 [tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (1)^{2}[\frac{1}{(7)^{2}} - \frac{1}{(1)^{2}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (0.02 - 1)[/tex]
= [tex]2.13 \times 10^{-18} J[/tex]
As, E = [tex]\frac{hc}{\lambda}[/tex]
Putting the given values and calculate the wavelength as follows.
E = [tex]\frac{hc}{\lambda}[/tex]
[tex]2.13 \times 10^{-18} J[/tex] = [tex]\frac{6.626 \times 10^{-34}Js \times 3 \times 10^{8 m/s}}{\lambda}[/tex]
[tex]\lambda = 9.33 \times 10^{-8}[/tex] m
= 93.3 nm
4). For [tex]n_{1}[/tex] = 7, [tex]n_{2}[/tex] = 6, Z for H = 1 [tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (1)^{2}[\frac{1}{(6)^{2}} - \frac{1}{(7)^{2}}][/tex]
= [tex]-2.178 \times 10^{-18} J \times (0.03 - 0.02)[/tex]
= [tex]-0.0217 \times 10^{-18} J[/tex]
As, E = [tex]\frac{hc}{\lambda}[/tex]
Putting the given values and calculate the wavelength as follows.
E = [tex]\frac{hc}{\lambda}[/tex]
[tex]0.0217 \times 10^{-18}J[/tex] = [tex]\frac{6.626 \times 10^{-34}Js \times 3 \times 10^{8 m/s}}{\lambda}[/tex]
[tex]\lambda = 916.03 \times 10^{-8}[/tex] m
= 9160.3 nm
Thus, we can conclude that transition from n = 7 to n = 6 will emit a photon with the longest wavelength.
