Question

Which photon energy could be absorbed by a hydrogen atom that is in the n 2 state?

Answer 1

The Rydberg equation describes the energy level transitions in the hydrogen atom only:

#bb(DeltaE = -R_H(1/n_f^2 - 1/n_i^2))#

where #R_H# is the Rydberg constant, and is actually the magnitude of the ground-state energy of the hydrogen atom, #|-"13.61 eV"| = "13.61 eV"#.

#n_f# and #n_i# are the final and initial values of the principal quantum number #n#, respectively.

Therefore, since going from #n = 1# to #n = 2# is the only possible transition upwards to #n = 2#:

#color(blue)(DeltaE) = -"13.61 eV"(1/(2)^2 - 1/(1)^2)#

#=# #color(blue)("10.21 eV")#

Or, in possibly more familiar units...

#color(blue)(DeltaE) = 10.21 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV") xx "1 kJ"/(1000 cancel"J") xx (6.0221413 xx 10^(23) "things")/"1 mol H atom"#

#=# #color(blue)("984.77 kJ/mol H atom")#

Answer 2

Given :

Hydrogen atom :

final state = $n_{2}$ = 2

Initial state = $n_{1}$ = 1

To Find :

Photon energy

$\Delta E$ = ? in transition

Explanation:

Rydberg equation describes the energy level transitions in the hydrogen atom

$\Delta E = R_{H} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$R_{H}$ is the Rydberg constant = 13.61eV

$\Delta E = 13.61eV \left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

      = 10.21eV

and we also know ;

$1eV = 1.602 * 10^{-19}$

and 1 mole of hydrogen have = $6.023 *10^{23}$ atoms .

Hence :

$\Delta E = 984.77 kJ/mol H atom$

#SPJ3