Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.57 nm. It then gives off a photon having a wavelength of 954.3 nm. What is the final state of the hydrogen atom? Values for physical constants can be found in g

Answer 1

Answer:

the final state of the hydrogen atom = 3

Explanation:

From the given information;

Let first calculate the amount of energy by the ground state atom during the atoms absorbs  photon light  by using the formula:

$E_{absorbs} = \dfrac{hc}{\lambda}$

where;

h = planck's constant = $6.626*10^{-34 }\ \ Js$

c = speed of light = $3.0*10^8 \ \ m/s$

λ = wavelength = 92.57 nm = 92.57 × 10⁻⁹ m

$E_{absorbs} = \dfrac{6.626*10^{-34 }\ \ Js * 3.0*10^8 \ \ m/s}{92.57*10^{-9} \ \ m}$

$E_{absorbs} = 2.15 *10^{-18} \ J$

The energy emitted by the hydrogen atom is calculated by using the same formula from above ; but here , the wavelength λ = 954.3 nm = 954.3 × 10⁻⁹ m

$E_{absorbs} = \dfrac{6.626*10^{-34 }\ \ Js * 3.0*10^8 \ \ m/s}{954.3*10^{-9} \ \ m}$

$E_{absorbs} = 2.08 *10^{-19} \ J$

The change in the energy absorbed is:

$\Delta E= 2.15 *10^{-18} \ J - 2.08 *10^{-19} \ J$

$\Delta E= 1.94 *10^{-18} \ J$

The final state of the atom can be determined by using the relation:

$\Delta E = R_H [\dfrac{1}{1^2}-\dfrac{1}{n^2_f}]$

where;

$R_H$ = Rydberg constant = 2.18 × 10⁻¹⁸ J

$\dfrac{\Delta E}{R_H} = [\dfrac{1}{1^2}-\dfrac{1}{n^2_f}] \\ \\ \\ \dfrac{1.94*10^{-18} \ J}{2.18*10^{-18} \ J } = [\dfrac{1}{1^2}-\dfrac{1}{n^2_f}] \\ \\ \\ 0.889 = [\dfrac{1}{1^2}-\dfrac{1}{n^2_f}] \\ \\ \\ 1 - 0.889 = \dfrac{1}{n^2_f} \\ \\ \\ 0.111= \dfrac{1}{n^2_f} \\ \\ \\ {n^2_f} = \dfrac{1}{0.111} \\ \\ \\ {n^2_f} = 9 \\ \\ \\ {n_f} = \sqrt{9} \\ \\ \\ \mathbf{n_f = 3}$

Thus; the final state of the hydrogen atom = 3