Question

Hello Everyone!
Out with some $\tt{Chemistry}$ questions today,
Topic - Atomic Structure :)
1. With increasing quantum number, the energy difference between adjacent energy levels in H atom decreases. Justify.
2. Calculate the wave number of shortest wavelength transition in Balmer series of atomic hydrogen.
3. Refer to Attachment.
$\longrightarrow$ This is the list from Min to Max Frequency. Where shall we keep the seismic waves in this list?
Give proper explanations :) ​

Answer 1

1 ) Electrons in the quantum model of an atom revolve in Bohr's orbit but are scattered into different orbitals. The centripetal force required for an electron to stay in the Bohr's orbit is provided by the electrostatic interaction between the electrons and the protons(in the nucleus). This electrostatic force follows the inverse-square relationship, that is, it varies as 1/r². As the principal quantum number increases, the electron enters into a farther orbit thus it's distance (r) from the nucleus (or the protons) increases and hence the magnitude of force decreases. Since, the magnitude of force reduces, the velocity of the electron also reduces, which implies that the Kinetic Energy of the orbit decreases.

Also, the total energy of an orbit, following inverse-square law, is equal in magnitude to its kinetic energy. That is TE = |KE|. Now, since KE got reduced, its TE (total energy) also decreases. And hence, the energy difference decreases with increase in quantum numbers.

2) The transition from 2nd orbit to the infinite orbit is what corresponds to the shortest wavelength in a Balmer series. Thus, its wave number (1/wavelength) is :

1/lambda = 1.097×10⁷(1/2² - 1/∞²) = 1.097×10⁷(1/4+1/∞) = 1.097×10⁷(1/4) = 0.27425×10⁷ = 2742500 m^-1.

Thus, the wave number of the shortest wavelength in Balmer series is 2742500 m^-1.

3) The minimum frequency (as in the attached image) corresponds to the Radio Waves which possess a frequency range of 10⁶ to 10⁸ Hz. Thus, the Seismic Waves tops the list with frequency ranging from 1Hz to 150Hz.

Answer 2

Answer:

1 energy difference between consecutive levels decreases as n increases. So, as the n increases, energy between the consecutive levels will decrease. Hence, with increasing quantum numbers the energy difference between adjacent levels in atoms decreases.

2 Rh =Rydberg constant=109737cm ^−1

The transition from n=2 to n=∞ will have shortest wavelength

wave number= v = 1/λ

=Rh ( 1/4 − 1/ ∞ )

= Rh/4

=27434.25cm^-1 (approx)

3 The minimum frequency corresponds to the ratio waves which processes a frequency range of 10^6 to 10^8 Hz .So the seismic waves tops the list with frequency ranging from 1Hz to 150Hz.