Answer:
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
Explanation:
From the question we are told that
Light wavelength [tex]\lambda_l=655nm[/tex]
Light wavelength atom fall [tex]x_L=5th-2nd[/tex]
Photon wavelength [tex]\lambda_p=435nm[/tex]
Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]
Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by
[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]
Therefore for initial drop of 5th to 2nd
[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]
Therefore for initial drop of 6th to 2nd
[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
Generally we subtract (5th to 2nd) from (6th to 2nd)
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]
Therefore for 6th to 5th stage is mathematically given by
[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]
[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
