Question

What are the frequency and wavelength
of a photon emitted during a transition
from n = 5 state to the n=2 state in the
hydrogen atom?​

Answer 1

Answer:

f = 4.3 * 10³³

λ = 6.9 * 10⁻²⁴

Explanation:intial orbit is 5 and the final is 2.

The speed of EM wave is c = 3 * 10⁸  m/s

Plank's constant h= 6.63 × 10⁻³⁴ Js

First, calculate E by using the following relation,

                 E = −13.6 [ 1/nf² - 1/ni⁵] eV

put the values and you'll get E= 2.856eV

now, frequency is f = E/h

f = 2.856/6.63 × 10⁻³⁴ = 4.3 * 10³³

now, wavelength is c = fλ

λ = c/f

λ = 3 * 10⁸/ 4.3 * 10³³ = 6.9 * 10⁻²⁴

Answer 2

Answer:

Hey buddy!!! Here's the answer:)

Explanation:

The frequency of any photoelectric phenomena can be determined when the energy is known, as Energy = Planck’s constant(h) * speed of light(c)/wavelength(&). The frequency is nothing but c/& as speed=frequency times wavelength. Thus, Energy = Planck’s constant(h)*frequency(v).

According to your case, the electron is in a transition state from n=5 to n=2. So, the total energy during the transition is given by E(net) = E(at state 5) - E(at state 2).

The energy corresponding to given state is determined by E(energy at nth state) = -13.6/(n^2). The 13.6 is the magnitude energy at the ground state or non-excited state. The negative sign implies that the electron is bound to nucleus and that by itself cannot come out of the atom, and additional energy is certainly need to excite or take it out of the atom.

Thus, the Energy at 5th state is -13.6/25 = -0.544 and the Energy at 2nd state is -13.6/4 = 3.4eV

E(net) = E5-E2 = 2.85 eV, and take a note that ALL THE ENERGY WHICH WE CALCULATE ARE IN ELECTRON VOLT (eV) AND NOT IN JOULES.

Thus, we know the net energy, and by substituting in energy formula (E=h*v) , we get v ~ 7*(10^14) hertz , where h = 6.6*(10^(-34)) joules seconds

Hope it helps. ALL THE BEST! :)