Answered

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Using the boh model of a He ion, what transition is most likelu to result in the emission of radiation with a wavelength of approximately 274 nm

Sagot :

okpalawalter8

Answer:

[tex]n=6\ to\ n=3[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=274 *10^{-9}m[/tex]

Bohr's constant [tex]R = 1.097 × 10^7 / m (or m−1)[/tex]

Helium atom [tex]z=2[/tex]

Generally the equation for Wavelength is mathematically given by

 [tex]\frac{1}{\lambda}=Rz^2(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]

 [tex]0.083=(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]

Therefore

The Range of n fall at

 [tex]n=6\ to\ n=3[/tex]

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