Question

Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun and other stars. What is the energy of a photon (in J) that is absorbed by He+ ions, when an electron is excited from the Bohr orbit with n = 1 to the n = 4 state? The energy of an electron in the nth level is

Answer 1

Answer : The energy of a photon absorbed by $He^+$ ions is $8.18\times 10^{-18}J$

Explanation :

The energy of an electron in the nth level is,

$E_n=-\frac{BZ^2}{n^2}$

where,

$E_n$ = energy of an electron in the nth level

B = constant = $2.18\times 10^{-18}J$

n = number of energy level

Z = charge on nucleus or number of protons

First we have to calculate the energy of an electron for n = 1 level.

$E_1=-\frac{BZ^2}{1^2}$

Z = charge on nucleus or number of protons for helium atom = 2

$E_1=-\frac{(2.18\times 10^{-18})\times (2)^2}{1^2}$

$E_1=-8.72\times 10^{-18}J$

Now we have to calculate the energy of an electron for n = 4 level.

$E_4=-\frac{BZ^2}{4^2}$

Z = charge on nucleus or number of protons for helium atom = 2

$E_4=-\frac{(2.18\times 10^{-18})\times (2)^2}{4^2}$

$E_4=-5.45\times 10^{-19}J$

Now we have to calculate the energy of a photon that is absorbed by $He^+$ ions.

$E=E_4-E_1$

$E=(-5.45\times 10^{-19})-(-8.72\times 10^{-18})$

$E=8.18\times 10^{-18}J$

Therefore, the energy of a photon absorbed by $He^+$ ions is $8.18\times 10^{-18}J$