(a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is initially in the third excited state.
What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?i) According to Bohr’s second postulate, we have 
But, as per De- broglie hypothesis
Therefore, 
; where is the de- broglie wavelength.
ii) Given, electron in the hydrogen atom is in the third excited state.
For third excited state, n = 4
For ground state n= 1
Now, total number of possible spectral lines is given by,
The transition states are as shown in the figure above.
The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm?
Which of these transitions correspond to emission of radiation of (i) maximum and (ii) minimum wavelength?
Given, wavelength of the photon, 
= 275 nm
Energy of the photon is given by, 
This corresponds to transition B as from the figure.
i) 
For maximum wavelength should be minimum.
Minimum energy corresponds to transition A.
ii) For minimum wavelength, should be maximum. Maximum energy corresponds to transition D.
In the study of Geiger-Marsdon experiment on scattering of α particles by a thin foil of gold, draw the trajectory of -particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study. From the relation R = Ro A1/3 where Ro is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
Consider an alpha-particle with initial K.E = 1/2 mv2 directed towards the center of nucleus of an atom.
The force that exists between nucleus and α-particle is Coulomb’s repulsive force. On account of this force, at the distance of closest approach ro , the particle stops and cannot go further closer to the nucleus. And, K.E gets converted to P.E.
That is, 
Hence, we can see that the size of the nucleus is approximately equal to the distance of closest approach, ro .
Now,
If ‘m’ is the average mass of the nucleon and r the radius of nucleus, then
Mass of nucleus = mA; where A is the mass number of the element.
Volume of the nucleus, V = 4/3 πR3
This implies, 
Therefore, the nuclear density is independent of mass number A.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy levels the hydrogen atoms would be excited?
Calculate the wavelengths of the first member of Lyman and first member of Balmer series.Amount of energy required to excite the electron = 12.5 eV
Energy of the electron in the nth state of an atom = 
; Z is the atomic number of the atom. [Z=1 for hydrogen atom]
Energy required to excite an atom from the initial state (ni) to the final state (nf) = 
This energy must be equal to or less than the energy of the incident electron beam.
… (1)
i=1 for ground state of hydrogen atom.
Energy of the electron in the ground state = −13.6 eV
Now, putting this in equation (1),
Since, the state cannot be a fractional number we have nf = 3
Therefore, hydrogen atom would be excited up to 3rd energy level.
b) Rydberg formula is given by, 
; 
is the wavelength and R is the Rydberg constant.
R = 1.097373157 × 10 7 m-1
For the first member of Lyman series, i=1; f=2
So, 
For the first member of the Balmer series, i=2; f=3
So, 
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