Question

Calculate the shortest wavelength of the electromagnetic radiation emitted by the hydrogen atom in undergoing a transition from the n = 6 level.

Answer 1

Answer:

$\lambda = 9.376*10^{-8} m$

Explanation:

From the question we are told that:

 $n=6level$

Generally the equation for Energy is mathematically given by

 $E=\frac{hc}{\lambda}$

Since

Energy difference will be maximum when electron return to ground state

And Shortest wavelength is emitted when  there is largest energy difference

Therefore

 $1/\lambda = R* (\frac{1}{nf^2} - \frac{1}{ni^2})$

Where

 $R=Rydberg\ constant$

 $R = 1.097*10^7$

Therefore

 $1/\lambda = 1.097*10^7* ((\frac{1}{1^2} - \frac{1}{6^2})$

  $\lambda = 9.376*10^{-8} m$