Question

nd (b) an electronic transition from the n 5 state to the n 6 state? What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen and (b) an electronic transition from the n 5 state to the n 6 state?

Answer 1

Answer:

a) $E_photon =0.306 eV$

b) $E_photon =0.166 eV$

Explanation:

The energy of the photon (E) for $n^th$ orbit of the hydrogen atom is given as:

$E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

where,

$E_o$ = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

$E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})$

$E_photon =0.306 eV$

b) Now for the transition from n = 5 to n = 6

$E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})$

$E_photon =0.166 eV$