Answer:
The wavelength is [tex]\lambda =97.3 nm[/tex]
Explanation:
Generally the series whose emission line show on the visible spectrum is the
Balmar series so this two emission line seen on the visible spectrum could either be due to the move of electron from
[tex]n=3 \to \ n=2[/tex]
OR
[tex]n=4 \to n=2[/tex]
This implies that the first excitement is from [tex]n_i=1 \to \ n_f=4[/tex]
So the energy change due to the excitement is mathematically represented as
[tex]\Delta E = R_H [\frac{1}{n_i^2} -\frac{1}{n_f^2} ][/tex]
substituting values
[tex]\Delta E= R_H [\frac{1}{1^2} -\frac{1}{4^2} ][/tex]
[tex]\Delta E= \frac{15}{16} R_H[/tex]
This energy change can also be represented as
[tex]\Delta E = \frac{hc}{\lambda}[/tex]
So [tex]\frac{15}{16} R_H = \frac{hc}{\lambda}[/tex]
=> [tex]\lambda = \frac{16hc}{15 R_H}[/tex]
Where [tex]R_H[/tex] is the Rydberg constant with a value of [tex]R_H = 2.18 * 10^{-18} J.[/tex]
h is the Planck's constant with values [tex]h = 6.626 * 10^{-34} m^2 kg / s[/tex]
c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{16(6.626 *10^{-34})(3*10^{8})}{15 * (2.18*10^{-18})}[/tex]
[tex]\lambda = 9.73 *10^{-8} \ m[/tex]
[tex]\lambda =97.3 nm[/tex]
