Q: What is the shortest wavelength present in the Paschen series of spectral lines?

Rydberg’s formula is given as hc/λ=21.76×10−19(1/n12−1/n22)h=6.6×10−34 is the plank's constant.c=3×108 is speed of life.The shortest wavelength present in the Paschen series of the spectral lines is given for values n1=3 and n2=∞.From the above values, λ=819nm

Q: 12.6.A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4th level. Detere the wavelength and frequency of photon.

For ground level, n1=1.E1=−13.67/n12=13.67eVThe atom is excited to higher level, n2=4.E2=−13.67/n22=13.67/16 eVE=E1−E2=hc/λλ=E1−E2hc⇒13.67−1613.676.6×10−34×3×108⇒λ=97nmc=λν⇒ν=3.1×1015Hz

Q: The radius of the innermost electron orbit of a hydrogen atom is 5.3×1011m. What are the radii of the n = 2 and n =3 orbits?

r1=5.3×10−11mLet r2 be the radius at n=2.r2=n2r1=2.12×10−10mSimilarly, for n=3,r3=n2r1=4.77×10−10m

Question 1

Choose the correct alternative from the clues given at the end of each statement:a) The size of the atom in the Thomsons model is ........... the atomic size in Rutherford's model. (much greater than/no different from/much less than.)b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomsons model/ Rutherfords model.)c) A classical atom based on .......... is doomed to collapse. (Thomsons model/ Rutherfords model.)d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomsons model/ Rutherfords model.)e) The positively charged part of the atom possesses most of the mass in .......... (Rutherfords model / both the models.)

Accepted Answer

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.(b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.(c) A classical atom based on Rutherford’s model is doomed to collapse.(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.(e) The positively charged part of the atom possesses most of the mass in both the models.

Question 2

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Accepted Answer

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67×10−27) is less than the mass of incident α−particles (6.64×10−27). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the  α−particles would not bounce back if solid hydrogen is used in the α−particle scattering experiment.

Question 3

What is the shortest wavelength present in the Paschen series of spectral lines?

Accepted Answer

Rydberg’s formula is given as hc/λ=21.76×10−19(1/n12​−1/n22​)h=6.6×10−34 is the plank's constant.c=3×108 is speed of life.The shortest wavelength present in the Paschen series of the spectral lines is given for values n1​=3 and n2​=∞.From the above values, λ=819nm

Question 4

12.4.A difference of 2.3eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Accepted Answer

Given: The difference in energy between the two energy levels is 2.3eV.To find: The frequency of radiation emitted when the atom makes a transition between the energy levels.The energy between the energy levels is E=2.3eV⇒2.3×1.6×10−19JThe frequency of the radiations emitted during transition is:E=hνν=Eh​⇒2.3×1.6×10−196.626×10−34​⇒5.5×1014Hz

Question 5

12.5.The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?

Accepted Answer

Ground state energy of hydrogen atom is E=−13.67eVThe kinetic energy of the electron is given as:KE=−E=+13.67eVThe potential energy of the electron is given as:PE=−2×KE=−27.2eV

Question 6

12.6.A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4th level. Detere the wavelength and frequency of photon.

Accepted Answer

For ground level, n1​=1.E1​=−13.67/n12​=13.67eVThe atom is excited to higher level, n2​=4.E2​=−13.67/n22​=13.67/16 eVE=E1​−E2​=hc/λλ=E1​−E2​hc​⇒13.67−1613.67​6.6×10−34×3×108​⇒λ=97nmc=λν⇒ν=3.1×1015Hz

Question 7

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels(b) Calculate the orbital period in each of these levels.

Accepted Answer

(a) v is the orbital speed of the electron in the hydrogen atom in level n.v=n1​4πϵo​(h/2π)e2​For n=1, →v1​=2.18×106m/sFor n=2, →v1​=1.09×106m/sFor n=3, →v1​=7.27×105m/s(b) Orbital perios is given by T=v2πr​;v=velocityRadius, r=πme2n2h2ϵo​​So,T1​=1.527×10−16, n=1 T2​=1.22×10−15, n=2 T3​=4.12×10−15, n=3

Question 8

The radius of the innermost electron orbit of a hydrogen atom is 5.3×1011m. What are the radii of the n = 2 and n =3 orbits?

Accepted Answer

r1​=5.3×10−11mLet r2​ be the radius at n=2.r2​=n2r1​=2.12×10−10mSimilarly, for n=3,r3​=n2r1​=4.77×10−10m

Question 9

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

Accepted Answer

Energy of the electron beam is =12.5eVWe know that, the energy of the electron in the nth state of an atoms is:E=n213.6z2​eVThe energy required to move the electron from one energy level to the other level is:E′=nf2​13.6​+n12​13.6​This energy must be equal to or less than the energy of the electron beam.−nf2​13.6​+ni2​13.6​=12.5 ___(i)Since for ground state:i=1Energy of the electron in the ground state =−13.6eVNow, putting this in equation (1),∴−nf2​13.6​+13.6=12.5⇒13.6−12.5=−nf2​13.6​⇒nf2​=1.113.6​=12.36⇒nf​=3.5Therefore, hydrogen atom would be excited up to 3rd energy level.

Question 10

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

Accepted Answer

r=1.5×1011m; v=3×104m/s;M is the mass of earth.According to Bohr model, Mvr=2πnh​.Thus, n=2.6×1074 is the quanta number that charaactrizes the earth revolution.

Question 11

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?

Accepted Answer

(a) The average angle of deflection of -particles by a thin gold foil predicted by Thomsons model is about the same than that predicted by Rutherfords mode(b) The probability of backward scattering (i.e., scattering of -particles at angles greater than 90) predicted by Thomsons model much less than that predicted by Rutherfords model.(c) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since, the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.(d) Thomson’s model.It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of α−particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Question 12

12.12.The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Accepted Answer

Radius of the first Bohr orbit is given by the relation,r1​=me​e24πϵ0​(2πh​)2​      ......(1)Where,ϵ0​= Permittivity of free spaceh= Planck's constantme​= Mass of an electrone= Charge of an electronmp​= Mass of a protonr= Distance between the electron and the protonCoulomb attraction between an electron and a proton is given as:FC​=4πϵ0​r2e2​     .....(2)Gravitational force of attraction between an electron and a proton is given as:FG​=r2Gmp​mc​​     .....(3)WhereG= Gravitation constant =6.67×10−11Nm2/kg2If an electrostatic (Coulomb) force and the gravitation force between an electron and a proton are equal, then we can write:FG​=FC​r2Gmp​mc​​=4πϵ0​r2e2​∴4πϵ0​e2​=Gmp​mc​     .....(4)Putting the value of equation ( 4 ) in equation ( 1 ), we get:r1​=Gmp​mc2​(2πh​)2​=6.67×10−11×1.67×10−27×(9.1×10−31)2(2×3.146.63×10−34​)2​≈1.21×1029mIt is known that the universe is 156 billion light years wide or 1.5×1027m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Question 13

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1​). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Accepted Answer

The charge of the muon m=207me​,Bohr radius is given as:re​=1/me​We know that for the first Bohr orbit re​=0.53×10−10mSo, at equilibrium, mr=me​re​The radius pf the muonic hydrogen atom r=2.56×−13mWe also know that, Ee​∝me​;  Ee​=−13.6eV⇒EEe​​=mme​​⇒E=−2.81 keV

Question 14

12.14.Classically, an electron can be in any orbit around the nucleus of an atom. Then what deteres the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ( 10−10m).(a) Construct a quantity with the dimensions of length from the fundamental constants e,me​,andc. Detere its numerical value.

Accepted Answer

According to Coulomb's law, the force between hydrogen nucleus and electron is giben as:F=4π ε0​1​.r2e2​⇒r=4π ε0​1​F.re2​Now, using the fundamental constants e,me​ and c, we will obtain a quantity which has the dimensions of length.We know that F.r  (force x distance) is the amount work or energy  and it is also given as =mc2∴r=4π ε0​1​mc2e2​=2.8×10−15m.It is comparatively smaller than the size of the atom.

Question 15

The total energy of an electron in the first excited state of the hydrogen atom is about −3.4eV.(a) What is the kinetic energy of the electron in this state?(b) What is the potential energy of the electron in this state(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Accepted Answer

In Bohrs model, mvr = nh and rmv2​=4πε0​r2Ze2​, which give,T=21​mv2=8πε0​rZe2​;r=Ze2m4πε0​h2​n2 These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity, we have V=−(Ze2/4πε0​r), which gives V=−2T and E=T+v=−T.(a) The quoted value of E = 3.4 eV is based on the customary choice of zero of potential energy at infinity. Using E = T, the kinetic energy of the electron in this state is + 3.4 eV.(b) Using V = 2T, potential energy of the electron is = 6.8 eV.(c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV independent of the choice of the zero of potential energy. The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen.

Question 16

If Bohrs quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

Accepted Answer

The quantization of the orbits of the planetary motions is not considered in the same manner as the Bohr's quantization postulates for the atoms because the value of the angular momentum in terms of the Plank's constant h will be of the order of 1070h. The high value of the angular momentum leads to the quantum levels of the order of 1070.The energy and the angular momentum for such high values of n will be comparatively very small which is not possible in case of the planets.

Question 17

Obtain the first Bohrs radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−) of mass about 207me​ orbits around a proton].

Accepted Answer

The charge of the muon m=207me​,Bohr radius is given as:re​=1/me​We know that for the first Bohr orbit re​=0.53×10−10mSo, at equilibrium, mr=me​re​The radius pf the muonic hydrogen atom r=2.56×−13mWe also know that, Ee​∝me​;  Ee​=−13.6eV⇒EEe​​=mme​​⇒E=−2.81 keV

Question 18

In the Rutherford's nuclear model of the atom, the nucleus (radius about 10−15 m ) is analogous to the sun about which the electron move in orbit (radius ≈10−10 m ) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth's orbit is about 1.5×1011 m. The radius of sun is taken as 7×108 m.

Accepted Answer

The ratio of the radius of electron's orbit to the radius of nucleus is (10−10 m)/(10−15 m)=105, that is, the radius of the electron's orbit is 105 times larger than the radius of nucleus. If the radius of the earth's orbit around the sun were 105 times larger than the radius of the sun, the radius of the earth's orbit would be 105×7×108 m= 7×1013 m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun.
It implies that an atom contains a much greater fraction of empty space than our solar system does.

Question 19

In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7MeVα-particle before it comes momentarily to rest and reverses its direction?

Accepted Answer

The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system's initial mechanical energy is Ei​, before the particle and nucleus interact, and it is equal to its mechanical energy Ef​ when the α-particle momentarily stops. The initial energy Ei​ is just the kinetic energy K of the incoming α - particle. The final energy Ef​ is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. (12.1).
Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy Ei​=Ef​ as
K=4πε0​1​d(2e)(Ze)​=4πε0​d2Ze2​
Thus the distance of closest approach d is given by
d=4πε0​K2Ze2​
The maximum kinetic energy found in α-particles of natural origin is 7.7MeV or 1.2×10−12 J. Since 1/4πε0​=9.0×109 N m2/C2. Therefore with e=1.6×10−19C, we have,
d=1.2×10−12 J(2)(9.0×109Nm2/C2)(1.6×​
=3.84×10−16Zm
The atomic number of foil material gold is Z=79, so that d(Au)=3.0×10−14 m=30fm.(1fm (i.e. fermi) =10−15 m.)
The radius of gold nucleus is, therefore, less than 3.0×10−14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the α-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Question 20

It is found experimentally that 13.6eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

Accepted Answer

Total energy of the electron in hydrogen atom is −13.6eV= −13.6×1.6×10−19 J=−2.2×10−18 J. Thus from Eq. (12.4), we have −8πε0​re2​=−2.2×10−18 J
This gives the orbital radius
r=−8πε0​Ee2​=−(2)(−2.2×10−18 J)(9×109 N m2/C2)(1.6×10−19C)2​
=5.3×10−11 m.
The velocity of the revolving electron can be computed from Eq. (12.3) with m=9.1×10−31 kg,
v=4πε0​mr​e​=2.2×106 m/s

Chapter 12 : Atoms