Question

A hydrogen atom in the ground state absorbs energy and makes a transition to the n = 3 state. The atom returns to the ground state by emitting two photons. What are the wavelengths of the two photons?
(A) 6.56 x 10-7m, 1.22 x 10-7m
(B) 6.56 x 10-7m, 5.45 x 10-7m
(C) 3.43 x 10-7m, 1.22 x 10-7m
(D) 5.45 x 10-7m, 3.43 x 10-7m

Answer 1

Answer:

at n= 3     λ = 656 nm

at n= 2     λ = 121.58 nm

Explanation:

Given details

transition of hydrogen atom from n = 2 to n = 3 state

Difference in energy between n = 3 state and n = 2 state :

      $= 2.18*10^{-18} \times [1/4 - 1/9] J = 3.03 \times 10^{-19} J$

    so, energy of photon is given as $=\frac{h\timesc}{\lambda}$

$E = 3.03*10^{-19}$ So solve for  wavelength

so, λ $= \frac{6.626*10^{-34}*3*10^8}{3.03*10^{-19}}m$

   =$6.56*10^{-7} m$

   = 656 nm

for second transition,

energy transmitted is given asΔE $=\frac{h\times c}{\lambda}$

and it is calculated as $= 2.18*10^{-18}*[1/1 -1/4] J$

E = 1.635*10-18 J  solving for wavelength in ENERGY equation we get

so, $\lambda' = 1.2158*10^{-7}$ m

      = 121.58 nm