When light shines through atomic hydrogen gas, it is seen that the gas absorbs light readily at a wavelength of 91.63 nm. what is the value of the principal quantum number n of the level to which the hydrogen is being excited by the absorption of light of this wavelength? assume that the most of the atoms in the gas are in the lowest level. (a) 14 (b) 16 (c) 11 (d) 21

D) 21

Explanation:

When gas absorbs light , electron at lower level jumps to higher level .

and the difference of energy of orbital is equal to energy of radiation absorbed.

Here energy absorbed is equivalent to wavelength of 91.63 nm

In terms of its energy in eV , its energy content is eual to

1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .

Electron is lying in lowest or first level ie n = 1.

Energy of first level

= - 13.6 / 1² = - 13.6 eV.

Energy of n th level = - 13.6 / n². Let in this level electron has been excited

Difference of energy

= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)

13.6 / n² = 13.6 - 13.57 = .03

n² = 13.6 / .03 = 453

n = 21 ( approx )

answer: (431,2)

step-by-step explanation:

the answer is b

explanation: