Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices

Answer 1

Complete Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices

Answer:

$n=2$

Explanation:

From the question we are told that:

Wavelength $\lambda=486nm=>486*10^{-9}$

Generally the equation for Atom Transition is mathematically given by

$\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where

Rydberg constant $R_{\infty}=1.097*10^7$

Therefore

$\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})$

$(\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}$

$n_1^2=3.98$

$n=1.99$

$n=2$