Question

A hydrogen atom is excited from n=1 t n=3 state the amount of energy absorbed by the atom will be

Answer 1

Answer:

$\boxed{\mathfrak{Amount \ of \ energy \ absorbed \ (\Delta E) = 2.489 \times 10^{-18} \ J}}$

Given:

$\rm n_i = 1 \\ \rm n_f = 3$

Explanation:

Amount of energy absorbed by the hydrogen atom:

$\bf \Delta E = R_H(\dfrac{1}{{n_i}^{2}} - \dfrac{1}{{n_f}^{2}} ) \: J$

$\rm R_H \longrightarrow$ Rydberg constant $(\rm 2.8 \times 10^{-18})$

$\rm \implies \Delta E = 2.8 \times {10}^{ - 18} (\dfrac{1}{{1}^{2}} - \dfrac{1}{{3}^{2}} ) \\ \\ \rm \implies \Delta E = 2.8 \times {10}^{ - 18} (1 - \dfrac{1}{9} ) \\ \\ \rm \implies \Delta E = 2.8 \times {10}^{ - 18} (\dfrac{9 - 1}{9} ) \\ \\ \rm \implies \Delta E = 2.8 \times {10}^{ - 18} \times \dfrac{8}{9} \\ \\ \rm \implies \Delta E = 2.489 \times {10}^{ - 18} \: J$

Answer 2

Energy to excite ground state to third excited state. 

3rd excited state  -> e- in [4th orbit or shell] 

so, (n1=1 , n2=4)

E = E1  - E4

= 13.6 [ 1/12 - 1/42] ev

= 13.6 [1-1/16]

= 13.6 [15/16] 

= 13.6x 0.9375 = [12.75ev]

E = 12.75 ev