Question

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

Answer 1

Answer:

$n=6\ to\ n=3$

Explanation:

From the question we are told that:

Wavelength $\lambda=274 *10^{-9}m$

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

Helium atom $z=2$

Generally the equation for Wavelength is mathematically given by

 $\frac{1}{\lambda}=Rz^2(\frac{1}{nf^2}-\frac{1}{nf^2})$

 $0.083=(\frac{1}{nf^2}-\frac{1}{nf^2})$

Therefore

The Range of n fall at

 $n=6\ to\ n=3$