Q: The first excited energy of a He+ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?

The energy of hydrogen ion is given byEn=−n2(13.6eV)Z2 For the first excited state (n=2), the energy of He+ion (with Z=2 ) will be −13.6eV. This is same as the ground state energy of a hydrogen atom.Similarly, for all the hydrogen like ions, the energy of the (n−1)th excited state will be same as the ground state energy of a hydrogen atom if Z=n.

Q: What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En=E1/n2 , orrn=a0n2 ?

Energy of nth state of hydrogen is given byEn=n2−13.6eV Energy of first excited state (n=2)of hydrogen, E1=4−13.6eV=−3.4eV This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10eV when the electron is widely separated from the proton so here our reference point energy is 10eV. Earlier The energy of first excited state was −3.4eV when the reference point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,E1′=−3.4eV−10eV=−13.4eV We still write En=E1/n2 , or rn=a0n2 because these formulas are independent of the refrence point enegy.

Q: We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?

When a photon of energy (E2–E1=hυ ) is incident on an atom in the ground state, the atom in the ground state E1 may absorb the photon and jump to a higher energy state (E2 ). This process is called stimulated absorption or induced absorption.Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon. We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.

Question 1

How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

Accepted Answer

Balmer series contains wavelengths ranging from 364 nm (for n2​=3 ) to 655 nm(n2​=∞). So, the given range of wavelength (380−780 nm)  lies in the Balmer series. The wavelength in the Balmer series can be found by​λ1​=R(221​−n21​) Here, R= Rydberg’s constant =1.097×107 m−1​ The wavelength for the transition from n=3 to n=2 is given by​λ1​1​=R(221​−321​)λ1​=656.3 nm​ The wavelength for the transition from n=4 to n=2 is given by​λ2​1​=R(221​−421​)λ2​=486.1 nm​ The wavelength for the transition from n=5 to n=2 is given by​λ3​1​=R(221​−521​)λ3​=434.0 nm​ The wavelength for the transition from n=6 to n=2 is given byλ4​1​λ4​​=R(221​−621​)=410.2 nm​ The wavelength for the transition from n=7 to n=2 is given by​λ5​1​=R(221​−721​)λ5​=397.0 nm​ Thus, the wavelengths emitted by the atomic hydrogen in visible range (380-780 nm ) are 5 .Lyman series contains wavelengths ranging from 91 nm  (for n2​=2) to 121  nm(n2​=∞). So, the wavelengths in the given range (50-100 nm) must lie in the Lyman series. The wavelength in the Lyman series can be found byλ1​=R(121​−n21​) The wavelength for the transition from n=2 to n=1 is given by​λ1​1​=R(121​−221​)λ1​=122 nm​ The wavelength for the transition from n=3 to n=1 is given by​λ2​1​=R(121​−221​)λ2​=103 nm​ The wavelength for the transition from n=4 to n=1 is given by​λ3​1​=R(121​−421​)λ3​=97.3 nm​ The wavelength for the transition from n=5 to n=1 is given by​λ4​1​=R(121​−521​)λ4​=95.0 nm​ The wavelength for the transition from n=6 to n=1 is given by​λ5​1​=R(121​−621​)λ5​=93.8 nm​ So, it can be noted that the number of wavelengths lying between 50  nm to 100  nm are 3 .

Question 2

The first excited energy of a He+  ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?

Accepted Answer

The energy of hydrogen ion is given byEn​=−n2(13.6eV)Z2​ For the first excited state (n=2), the energy of He+ion (with Z=2 ) will be −13.6eV. This is same as the ground state energy of a hydrogen atom.Similarly, for all the hydrogen like ions, the energy of the (n−1)th excited state will be same as the ground state energy of a hydrogen atom if Z=n.

Question 3

Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?

Accepted Answer

As the electron collides, it transfers all its energy to the hydrogen atom.The excitation energy to raise the electron from the ground state to the nth state is given by: E=(13.6eV)×(121​−n21​) Substituting n=2, we getE=10.2eV Substituting n=3, we getE=12.08eV Thus, the atom will be raised to the second excited energy level.So, when it comes to the ground state, there is transitions from n=3 to n=1.Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).

Question 4

When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in Lyman series only. Explain.

Accepted Answer

White radiations are x-rays that have energy ranging between 5-10 eV.When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed only in the Lyman series.At room temperature, almost all the atoms are in ground state.The minimum energy required for absorption is 10.2eV (for a transition from n=1 to n=2 ). The white radiation has photon radiations that have an energy of around 10.2eV. So, they are just sufficient to transmit an electron from n=1 to n=2 level. Hence, the absorption lines are observed only in the Lyman series.

Question 5

Balmer series was observed and analysed before the other series. Can you suggest a reason for such an order?

Accepted Answer

The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364 nm (for n2​  = ∞ to 655 nm (for n2​  =3).

Question 6

What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En​=E1​/n2 , orrn​=a0​n2 ?

Accepted Answer

Energy of nth  state of hydrogen is given byEn​=n2−13.6​eV Energy of first excited state (n=2)of hydrogen, E1​=4−13.6​eV=−3.4eV This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10eV  when the electron is widely separated from the proton so here our reference point energy is 10eV. Earlier The energy of first excited state was −3.4eV  when the reference point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,E1′​=−3.4eV−10eV=−13.4eV We still write En​=E1​/n2 , or rn​=a0​n2  because these formulas are independent of the refrence point enegy.

Question 7

The difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series. Explain.

Accepted Answer

The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).The frequency of the radiation emitted for transition from n1​ to n2​ is given by f=k(n12​1​−n22​1​)Here,k is a constant.For the series limit of Lyman series,​n1​=1n2​=∞​ Frequency, f1​=k(121​−∞1​)=k For the first line of Lyman series,​n1​=1n2​=2​ Frequency, f2​=k(121​−221​)=43k​ For series limit of Balmer series,​n1​=2n2​=∞​ f1​=k(221​−∞1​)=4k​ f1​−f3​=f2​ Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.

Question 8

The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?

Accepted Answer

The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.1 eV = 1.6 × 10–19  JThe numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.

Question 9

We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?

Accepted Answer

When a photon of energy (E2​–E1​=hυ ) is incident on an atom in the ground state, the atom in the ground state E1​  may absorb the photon and jump to a higher energy state (E2​ ). This process is called stimulated absorption or induced absorption.Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon. We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.

Question 10

An atom is in its excited state. Does the probability of its coming to ground state depend on whether the radiation is already present or not? If yes, does it also depend on the wavelength of the radiation present?

Accepted Answer

When an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.Ratio of the coefficient for stimulated transition to spontaneous transition is given by:R=Bρ(ν)A​=ekThν​−1 For microwave region​ν=1010( say )R=e1.38×10−23×3006.6×10−34×1010​−1=e0.0016−1=0.0016​ This implies that stimulated transition dominate in this region.For visible region​ν=1015⇒R=e160−1⇒R>>1​ So here spontaneous transition dominate.

Question 11

The minimum orbital angular momentum of the electron in a hydrogen atom is:

Accepted Answer

h/2π According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplte of h/2π ∴Ln​=2πnh​ Here,n= Principal quantum numberThe minimum value of n is 1 .Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by L=2πh​

Question 12

Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV and 1.9 eV. These photons must come from:

Accepted Answer

Either two atoms or three atomsThe energies of the photons emitted can be expressed as follows:13.6(121​−221​)eV=10.2eV 13.6(121​−321​)eV=12.1eV 13.6(221​−321​)eV=1.9eV The following table gives the transition corresponding to the energy of the photon:A hydrogen atom consists of only one electron. An electron can have transitions, like from n=3 to n=2 or from n=2 to n=1, at a time.So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has n=3 to n=2 and then n=2 to n=1 electronic transition and the other has n=3 to n=1 electronic transition).

Question 13

Suppose, the electron in a hydrogen atom makes transition from n = 3 to n = 2 in 10−8  s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is:

Accepted Answer

10−42 N−m The angular momentum of the electron for the nth state is given by Ln​=2πn h​ Angular momentum of the electron for n=3,Li​=2π3h​ Angular momentum of the electron for n=2,Lf​=2π2h​ The torque is the time rate of change of the angular momentum.Torque, τ=tLf​−Li​​ ​=10−8(2 h/2π)−(3 h/2π)​=10−8−(h/2π)​=10−8−10−34​[∵2πh​≈10−34 J−s]=−10−42 N−m​ The magnitude of the torque is 10−42 N-m.

Question 14

In which of the following transitions will the wavelength be minimum?

Accepted Answer

n=2 to n=1For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated byλ1​=RZ2(n1​1​−n2​1​) Here, R is the Rydberg constant.For the transition from n=5 to n=4, the wavelength is given by​λ1​=RZ2(421​−521​)λ=9RZ2400​​ For the transition from n=4 to n=3, the wavelength is given by λ1​=RZ2(321​−421​)   λ=7RZ2144​ For the transition from n=3 to n=2, the wavelength is given by frac1λ=RZ2(221​−321​)  λ=5RZ236​ For the transition from n=2 to n=1, the wavelength is given by λ1​=RZ2(121​−221​)  λ=RZ22​ From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n=2 to n=1 will be minimum.

Question 15

In which of the following systems will the radius of the first orbit (n = 1) be minimum?

Accepted Answer

Doubly ionized lithiumFor a hydrogen-like ion with Z protons in the nucleus, the radius of the nth state is given byrn​=Zn2a0​​ Here, a0​=0.53pm For lithium,Z=3Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

Question 16

In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?

Accepted Answer

Doubly ionized lithiumThe wavelength corresponding the transition from n2​ to n1​ is given by:λ1​=RZ2(n12​1​−n22​1​) Here,R= Rydberg constantZ= Atomic number of the ionFrom the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.Therefore, the wavelength corresponding to n=2 to n=1 will be minimum in doubly ionized lithium ion because for lithium, Z=3.

Question 17

Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of principal quantum number n?

Accepted Answer

(c)The speed (v) of electron can be expressed asv=2∈0​hnZe2​….(1) Here,Z = Number of protons in the nucleuse= Magnitude of charge on electron chargen= Principal quantum numberh= Planck's constantIt can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number (n).Therefore, the graph between them must be a rectangular hyperbola.The correct curve is (c).

Question 18

The energy of an atom (or ion) in its ground state is -54⋅4eV. It may be:

Accepted Answer

He+ The total energy of a hydrogen-like ion, having Z protons in its nucleus, is given by: E=−n213.6Z2​eV Here, n= Principal quantum numberFor ground state,n=1∴  Total energy, E=-13.6 Z2eV For hydrogen,Z=1∴ Total energy, E=-13.6 eV For deuterium,z=1: Total energv, E=-13.6 eV For He+ Z=2∴  Total energy, E=-13.6 ×22=−54.4eV For Li++  Z=3∴  Total energy, E=-13.6×32=−122.4eV Hence, the ion having an energy of −54.4eV  in its ground state may be He+ .

Question 19

As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom:

Accepted Answer

IncreasesThe electric potential energy of hydrogen atom with electron at the nth state is given by:V=−n22×13.6​ As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.

Question 20

The radius of the shortest orbit in a one-electron system is 18 pm. It may be:

Accepted Answer

Li++ The radius of the nth orbit in one electron system is given by rn​=Zn2a0​​ Here, a0​=53pm For the shortest orbit,n=1For hydrogen,Z=1∴ Radius of the first state of hydrogen atom =53pm For deuterium,Z=1∴ Radius of the first state of deuterium atom =53pm For He+  Z=2∴ Radius of He+ atom =253​pm=26.5pm For Li++ Z=3∴ Radius of Li++ atom =353​pm=17.66pm≈18pm The given one-electron system having radius of the shortest orbit to be 18pm  may be Li++ .

Chapter 21 : Bohr'a Model and Physics of the Atom