Question

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=4 to n=1.energy emitted:

Answer 1

Answer:

$\mathbf{\Delta E = 2.042 \times 10^{-19} \ J}$

Explanation:

Using the Rydberg expression to calculate the change in energy of a photon emitted, we have :

$\Delta E = R ( \dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

where;

R = Rydberg constant = $2.178 \times 10 {-18} \ J$

$n_f$ = 1    and $n_i$ = 4

replacing our values into equation, we get:

$\Delta E = 2.178 \times 10^{-19} (\dfrac{1}{1^2}-\dfrac{1}{4^2})$

$\Delta E = 2.178 \times 10^{-19} (\dfrac{1}{1}-\dfrac{1}{16})$

$\Delta E = 2.178 \times 10^{-19} (\dfrac{15}{16})$

$\mathbf{\Delta E = 2.042 \times 10^{-19} \ J}$