In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10^11 m with orbital speed 3 × 10^4 m/s.

Q: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution
around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s.(Mass of earth = 6.0 × 1024 kg.)

Sol: Given, radius of orbit r = 1.5 × 1011 m

Orbital speed v = 3 × 104 m/s ;

Mass of earth M= 6 × 1024 kg

Angular momentum, mvr = nh/(2 π)

or, n = 2πvrm/h ;

[where, n is the quantum number of the orbit]

= (2 × 3.14 × 3 × 104 × 1.5 × 1011 × 6 × 1024 )/(6.63 × 10-34 )

= 2.57 × 1074

or, n = 2.6 × 1074.

Thus,the quantum number is 2.6 × 1074

Thus, the quantum number is 2.6 × 1074 which is too large.

The electron would jump from n = 1 to n = 3

E3 = (-13.6)/32 = -1.5 eV.

So, they belong to Lyman series.

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted ?

Q: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of
wavelength will be emitted ?

Sol: Energy of electron beam E = 12.5 eV

= 12.5 × 1.6 × 10-10 J

Planck’s constant h = 6.63 × 10-34 j-s

Velocity of light c = 3 × 108 m/s

Using the relation E = hc/λ , λ = (6.62 × 10-34 × 3 × 108)/(12.5 ×1.6 × 10-19 )

= 0.993 × 10-7 m = 993 × 10-10 m = 993 A°

This wavelength falls in the range of Lyman series (912 A° to 1216 A°)

Thus, we conclude that Lyman series of wavelength 993 A° is emitted.

A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. What is the initial state of hydrogen atom ?

Q: A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy
of 10.2 eV.
(i) What is the initial state of hydrogen atom?
(ii) What is the final state of hydrogen atom?
(iii) What is the wavelength of the photon emitted ?

Sol: (i) Let n1 be initial state of electron.

Then E1 = -13.6/(n12 ) eV ; Here E1 = -0.85 ev,

Therefore , -0.85 = 13.6/(n12 )

or, n1 = 4.

(ii) Let n2 be the final excitation state of the electron. Since excitation energy is always measured with
respect to the ground state,

Therefore , ΔE = 13.6 [1- 1/(n22 )]

here , ΔE = 10.2 eV, therefore, 10.2 = 13.6 [1 – 1/(n22 )]

or, n2 = 2

Thus the electron jumps from n1 = 4 to n2 = 2.

(iii) The wavelength of the photon emitted for a transition between n1 = 4 to n2 = 2,is given by

1/( λ) = R [1/(n22 )– 1/(n12 )]

(or) 1/λ=1.09 × 107 [1/22 – 1/42 ]= 4860 A ̇.

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^(-11) m. what are the radii of the n = 2 and n = 3 orbits ?

Q: The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. what are the radii of
the n = 2 and n = 3 orbits ?

Sol: Given, the radius of the innermost electron orbit of a hydrogen r1 = 5.3 × 10-11 m

As we know that rn = n2 r1

For n = 2, radius r2 = 22 r1

r2 = 4 × 5.3 × 10^(-11) = 2.12 × 10-10 m.

For n = 3, radius r3 = 32 r1

r3 = 9 × 5.3 × 10-11 = 4.77 × 10-10 m.

A doubly ionized lithium atom is hydrogen like with atomic number Z = 3. Find the wavelength of the radiation required to excite the electron in Li^(2+) form the first to the third Bohr orbit.

Q: A doubly ionized lithium atom is hydrogen like with atomic number Z = 3. Find the wavelength of the
radiation required to excite the electron in Li2+ form the first to the third Bohr orbit. Given the
ionization energy of hydrogen atom as 13.6 eV.

Sol: The energy of nth orbit of a hydrogen- like atom is given as En = -(13.6 Z2)/n2 Thus for Li2+ from, as Z = 3,

The electron energies of the first and third Bohr orbits are for n = 1, E1 =-122.4 eV, for
n = 3, E3 = -13.6 eV.

Thus the energy required to transfer an electron from E1 level to E3 level is,

E= E3– E1 = -13.6-(-122.4) = 108.8 eV.

Therefore, the radiation needed to causes this transition should have photons of this energy. hv = 108.8 eV.

The wavelength of this radiation is  , λ = hc/(108.8 eV)= 114.25 A°

A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting photons of energies 10.20 eV respectively….

Q: A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting photons of energies 10.20 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV )

Sol: The electronic transitions in a hydrogen like atom from a state n_2 to a lower state n_1 are given by

$\large \Delta E = 13.6 Z^2(\frac{1}{n_1^2} – \frac{1}{n_2^2} ) $ for the transition from a higher state n to the first excited state n_1 = 2, the total energy released is (10.2 + 17.0) eV or 27.2 eV

Thus ΔE = 27.2 eV , n_1 = 2 and n_2 = n.

We have , $\large 27.2 = 13.6 Z^2(\frac{1}{4} – \frac{1}{n^2} ) $ ….(i)

For the eventual transition to the second excited state n_1 = 3, the total energy released is (4.25 + 5.95)eV or 10.2 eV.

Thus , $\large 10.2 = 13.6 Z^2(\frac{1}{9} – \frac{1}{n^2} ) $ …(ii)

Dividing the eq. (i) by Eq. (ii) & solving we get ,

n2 = 36 ; or, n = 6

Substituting n = 6 in any one of the above equations, we obtain Z2 = 9 (or) Z = 3,

Thus n = 6 and Z = 3.

Consider a hydrogen-like atom whose energy in n^th excited state is given by E_n = – 13.6Z^2/n^2 when this excited atom makes transition from excited state have energy transition from excited state to ground state…

Q: Consider a hydrogen-like atom whose energy in nth excited state is given by $\large E_n = -\frac{13.6 Z^2}{n^2} $ when this excited atom makes transition from excited state have energy transition from excited state to ground state most energetic photons have energy Emax = 52.224 eV and least energetic photons have energy Emin = 52.224 eV. Find the atomic number of atom the state of excitation.

Sol: Maximum energy is liberated for transition En → 1 and minimum energy for En → En-1

Hence, $\large \frac{E_1}{n^2} – E_1 = 52.224 eV $ …(i)

And , $\large \frac{E_1}{n^2} – \frac{E_1}{(n-1)^2} = 1.224 eV $ …(ii)

Solving above equations simultaneously, we get

E1 = -54.4 eV and n = 5 ; Now E1=-(13.6 Z2)/12 = -54.4 eV

Hence, Z = 2 i.e., gas in helium originally excited to n = 5 energy state.