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**NCERT Class 11 ****Chemistry** Chapter 2 **Structure of Atom**

**Chemistry**Chapter 2

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**Structure of Atom**

**Structure of Atom****Chapter: 2**

Part – I |

**1. (i) Calculate the number of electrons which will together weigh one gram. **

Ans: The number of electrons which will weigh 1 g is 9.11 × 10^{-31} kg

= 9.11 × 10^{-31} kg = 1 electrones

∴ number of electrons that will weigh 1g (1 × 10^{-3} kg)

1/ 9.11 × 10^{-31} × 10^{-3 } electrones

= 1.1097 × 10^{-3 }electrones.

**(ii) Calculate the mass and charge of one mole of electrons.**

Ans: **Mass of one electron: **

= (9.11 × 10^{-31}) × (6.022 × 10^{23})

= 54.8604 × 10^{-7} Kg

**Charge on one electron:**

= 1.602 × 10^{-19} × 6.022 × 10^{23}

= 9.647 × 10^{4} coulombs.

**2. (i) Calculate the total number of electrons present in one mole of methane. **

Ans: Atomic number of carbon = 6

Number of electrons in carbon = 6

Atomic number of hydrogen = 1

Number of electrons in each hydrogen atom = 1

Total electrons in one molecule of methane = 6 + 4= 10

1 molecule of 𝐶𝐻_{4} contains electrons = 10

Number of electrons in one mole of methane = 10 × 6.023 × 10^{23} electrons.

= 6.023 × 10^{24} electrons.

**(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of **^{14}**C. (Assume that mass of a neutron = 1.675 × 10**^{–27}** kg).**

Ans: Mass of ^{14} C = 7 mg= 7 × 10-3

Molar mass of ^{14}C = 14 g/mol

∴ Amount of ^{14}C in 7 mg Sample

= 7 × 10-3g / 4g / mol

= 5 × 10^{-4} mol

Each carbon -14 nucleus contains 8 neutrons, hence

(a) The number of neutrons present in 7 mg (7 × 10-3g) of C-14

= 7 × 10-3g × 103 / 6.02 × 1023 × 8

= 2.408 × 1021 neutrons

(b) Total mass of neutrons in 7 mg sample of 14C

= 2.41 × 10^{21} × 10-^{27} kg

4.0347 × 10^{-6}

**(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH**_{3}** at STP. Will the answer change if the temperature and pressure are changed?**

Ans: Mass of NH_{3} = 34mg = 3410^{ -3}g

Molar mass of NH = 17 g / mol

Amount of ammonia = (34 × 10^{–}3 g) / 17 g / mol

= 2 × 10^{-3} mol

(a) Number of protons present in 1 molecule of ammonia = 7 + 3 = 10

Number of molecules present in 34 mg of ammonia is

= (6.022 × 10^{2}3 × 32 × 10^{-3}) / 17

= 10 × 1.2044 × 10^{21}

= 1.2044 × 10^{22} protons.

(b) Mass of proton = 1.6726 × 10^{−27 }kg.

There, total mass of protons in 34 mg of

NH3 = 1.2046 × 10^{22 }× 1.675 × 10^{-27} kg

**3. How many neutrons and protons are there in the following nuclei? **_{6}^{13}**C, **_{8}^{16 }**O, **_{12}^{24 }**Mg, **_{20}^{26}**Fe, **_{38}^{88}**Sr.**

Ans:

Nucleus | Mass number, A | Atomic number, Z | Number of protons | Number of neutrons= A-Z |

_{6}^{13}C | 13 | 6 | 6 | 13 – 6 = 7 |

_{8}^{16 }O | 16 | 8 | 8 | 16 – 8 = 8 |

_{12}^{24 }Mg | 24 | 12 | 12 | 24 – 12 = 12 |

_{20}^{26}Fe | 56 | 26 | 26 | 56 – 26 = 30 |

_{38}^{88}Sr | 88 | 38 | 38 | 88 – 38 = 50 |

**4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)**

**(i) Z = 17, A = 35.**

Ans: _{17}^{35}Cl.

**(ii) Z = 92, A = 233. **

Ans: ^{ 233}_{92}U.

**(iii) Z = 4, A = 9. **

Ans: ^{9}_{4}Be.

**5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ū ) of the yellow light.**

Ans: v = c/λ

V is frequency

C is the speed of light (approximately 3 × 10^{8} m/s)

λ is wavelength.

Ū = 1 / λ

λ = 580 nm = 580 × 10^{-19} m

Frequency, v = c/λ = 3.0 × 10^{8} ms -1 / 580 × 10^{-9} m

= 3.0 × 10^{8} / 5.80 × 10^{-7} m

= 5.17 × 10^{14} s^{-1}

Wave number, ū = 1 / λ = 1 / 580 × 10^{-9} m

= 1 / 5.8 × 10^{-7}

= 1.72 × 10^{6} m^{-1.}

**6. Find energy of each of the photons which:**

**(i) correspond to light of frequency 3 ×10**^{15}** Hz.**

Ans: E is the energy of the photon,

E = hv

hv = (6.626 × 10^{−34} J⋅s)

E = hv = (6.626 × 10^{−34} J⋅s) (3 × 10^{15} s-1)

= 19.88 × 10^{-19}J

=1.988 × 10^{-18}J

**(ii) have wavelength of 0.50 Å. **

Ans: The energy of photon is E = hc / λ

**First, convert the wavelength from angstroms to metres:**

λ = 0.50 Å = 0.50 ×10^{−10 }m

E = hc / λ

= (6.626 × 10^{−34} J⋅s) (3 × 10^{15} s^{-1}) / 0.50 × 10^{−10 }m

E = 3.98 × 10^{-15}s

**7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10**^{–10}** s.**

Ans: Calculate Wavenumber, 1/ period

= 1 / 2.0 × 10^{–10} s.

= 5 × 10^{9}s^{1.}

Calculate Wavenumber, (λ) = c /v

= (3 × 10^{8} ms-1) / ( 5 × 10^{9} s^{-1})

= 6.0 × 10^{-2} m.

Calculate Wavenumber, (Ū) = 1 / λ

= 1 / 6 × 10-2m

= 16.66m-1.

**8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?**

Ans: Wavelength (λ) = 4000 pm

= 4000 × 10^{-12} m

= 4 × 10^{-9}

∴ n = E × λ / h × c

E = 1 J

λ = 4 × 10^{-9} m

h = 6.626 × 10^{-34}

c = 3 × 10^{8} m-1

= (1 J) × (4 × 10^{-9} m) / (6.626 × 10^{-34}) (3 × 10^{8} m-1)

= 2.012 × 10^{16} photons.

**9. A photon of wavelength 4 × 10**^{–7}** m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate.**

**(i) the energy of the photon (eV).**

Ans: Energy of photon (E)

(E) = hv = hc / λ

= λ = 4 × 10−7 m

= h = 6.626

= c = 3 × 108 ms-1

= hc / λ = (6.626 × 10^{-34} js) × (3 × 10^{8} ms-1)/4 × 10^{-7} m

= 4.97 × 10^{-19} J

= 4.97 × 10^{-19} / 1.602 × 10^{-19}

eV = 3.10

**(ii) the kinetic energy of the emission. and**

Ans: The kinetic energy (KE) pf an emitted photoelectron is

= (½ mv^{2})

= hv – hv_{0}

= 3.1 – 2.13 = 0.97eV

**(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).**

Ans: 1/2 mv2 = 0.97 eVo

= 0.97 × 1.602 × 10^{-19} J

= 3.10 – 2.13

= 0.97 eV.

∵ 1/2 (9.11 × 10^{-31 }kg ) × v2

= 0.97 × 1.6 × 10-19 j

= v^{2} = 0.341 × 10^{12 }

= 34.1 × 10^{10}

V = 5.84 × 105 ms^{-1}

**10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol**^{–1}**.**

Ans: Wavelength of radiation,

E = hc / λ

**Where:**

h = Planck’s constant = 6.626 × 10^{−34} J⋅s

c = Speed of light = 3 × 108 m/s

λ = Wavelength = 242 nm

= 242 × 10−9 m

Energy (per mole) of the photons

N_{A}. hc / λ

= 6.023 × 1023 mol^{-1} × 6.626 × 10^{-24} js × 3 × 108 ms^{-1}/ 242 × 10^{-9} m

= 4.95 × 105 J mol ^{-1}

= 494 kJ mol^{-1}.

**11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.**

Ans: Power of bulb, P = 25

Watt (w) = 25 Js-1

Energy of one photon, E =

The energy of one photon is

∴ rate of emission

= 25 Js-1 / 3.487 × 10^{-19} J

= 7.18 × 10^{-19} s^{-1}.

**12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.**

Ans: The threshold frequency is the minimum frequency of radiation required to eject electrons from a metal surface. It is equal to the frequency of the incident radiation when the kinetic energy is zero.

= 6800 Å = 6800 × 10-10 m

Because c = vλ

**13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?**

Ans:

= 20564.4 cm^{-1}

**Rearranging to solve for wavelength λ lambdaλ: **

λ= 1/v = 1 / 20564.4cm^{-1}

= 486 × 10^{-7}cm

= 486 × 10^{-9}m

= 486 nm.

The colour corresponding to this wavelength.

**14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionisation enthalpy of H atom (energy required to remove the electron from n =1 orbit).**

Ans: The energy of an election in nth orbit is

For ionisation from 5th orbit, n^{1} = 5, n^{2} = ∞

For ionisation from 1st orbit, n^{1} = 1, n^{2} = ∞

Hence,it requires significantly less energy to ionise the hydrogen atom from the n = 5 orbit than from the n = 1 orbit.

**15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?**

Ans: Number of lines produced when electron from nth drops to ground state.

**Number of emission lines:** Σn(n−1)/2

= Σ (6-1) = 5

= 5 + 4 + 3 + 2 + 1 = 15

Thus, the maximum number of emission lines is 15.

**16. (i) The energy associated with the first orbit in the hydrogen atom is – 2.18 × 10**^{–18}** J atom**^{–1}**. What is the energy associated with the fifth orbit?**

Ans: The expression for the energy associated with nth orbit in hydrogen atom is

**(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.**

Ans: Radium of Bohar’s n^{th} orbit for hydrogen atom is given by,

r_{n = } 0.529nm × n^{2}

n = 5

r_{5} = 5.029 × 5^{2} nm

= 1.3225nm.

**17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.**

Ans: According to Balmer formula

For λ to be longest ⊽ should minimum. this can be so when n_{2} is minimum

i.e., n^{2} = 3

∴ Wavenumbe r(⊽) = 1.523 × 10^{6}m^{-1}

**18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is – 2.18 × 10**^{–11}** ergs.**

Ans: Energy to shift electron from 1st to 5th orbit Energy of an electron in the nth Bohr orbit is given by

Energy required to go to the 5th orbit

ΔE = E_{5} – E_{1}

= 8.72 × 10^{-15} ergs- (-218 × 10-13 erg)

= 2.08 × 10^{-11} ergs

For wavelength,

λ = 9.511 × 10-6 cm

= 951.1 × 10-8 cm

= 956 Å

∴ The wavelength of emitted radiation will be 956 Å.

**19. The electron energy in hydrogen atom is given by E**_{n}** = (–2.18 × 10**^{–18}**)/n**^{2}** J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?**

Ans: **Calculate the energy of the electron in the n = 2 orbit:**

**The energy for the n-th orbit is given by:**

Energy required to remove an electron from n = 2

= 2.18 × 10^{-18}J

= 5.42 × 10^{-19} J

**The relationship between energy and wavelength is given by:**

λ = 3.647 × 10-7 m

= 3.647 × 10-5 cm

∴ The wavelength of the light that can be used to cause this transition is 3.67 × 10-5 cm.

**20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 10**^{7}** ms**^{–1}**.**

Ans: **To calculate the wavelength of an electron moving with a given velocity, we use the de Broglie wavelength formula:**

λ = h / mv

λ is the de Broglie wavelength

V (velocity of electron) = 2.05 × 10^{7} ms^{-1}

M (mass of the electron) = 9.109 × 10^{−31} kg

h (Planck’s constant) = 6.626 × 10^{−34}J⋅s

= 3.53 × 10^{-11} m.

**21. The mass of an electron is 9.1 × 10**^{–31}** kg. If its K.E. is 3.0 × 10**^{–25}** J, calculate its wavelength.**

Ans: **The kinetic energy of the electron is given by:**

= 812 ms^{-1}

**The de Broglie wavelength λ is given by:**

= 8.967 × 10-7 m

= 8967 Å.

**22. Which of the following are isoelectronic species i.e., those having the same number of electrons?**

**Na**^{+}**, K**^{+}**, Mg2**^{+}**, Ca2**^{+}**, S2**^{–}**, Ar.**

Ans: **Number of electrons are:**

Na^{+} | 11-1 | 10 |

K^{+} | 19-1 | 18 |

Mg2^{+} | 12-2 | 10 |

Ca2^{+} | 20-2 | 18 |

S2^{–} | 16+2 | 18 |

Ar | 18 |

**So, the isoelectronic species are:**

Na^{+}, Mg2^{+}, (both have 10 electrons)

K^{+},^{,}Ca2^{+}, S2^{–}, Ar. (all have 18 electrons).

**23. (i) Write the electronic configurations of the following ions: (a) H**^{–}** (b) Na**^{+}** (c) O**^{2–}** (d) F**^{–}

Ans: **The electronic configuration of the given species are: **

(a) | H^{–} | 1s^{2} |

(b) | Na^{+} | 1s^{2 }2s^{2} 2p^{6 }3s^{1} |

(c) | O^{2} | 1s^{2 }2s^{2} 2p^{6} |

(d) | F^{–} | 1s^{2 }2s^{2} 2p^{6} |

**(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s**^{1}** (b) 2p**^{3}** and (c) 3p**^{5}**?**

Ans:

Special | Outer electronicconfiguration | Complete Electronicconfiguration | Total no. of electrons | Atomic number |

(a) | 2s^{1} | 1s^{2 }2s^{2} 2p^{6 }3s^{1} | 11 | 11 |

(b) | 2p^{3} | 1s^{2 }2s^{2} 2p^{3} | 7 | 7 |

(c) | 3d^{6} | 1s^{2 }2s^{2} 2p^{6 }3p^{6}4s^{2}3d^{6} | 26 | 26 |

**(iii) Which atoms are indicated by the following configurations?**

(a) [He] 2s^{1}

(b) [Ne] 3s^{2} 3p^{3}

(c) [Ar] 4s^{2} 3d^{1}

Ans:

Outer electronicconfiguration | Complete Electronicconfiguration | Total no. of electrons | Atomic number | Element |

2s^{1} | 1s^{2 }2s^{1} | 3 | 3 | Li |

2p^{3} | 1s^{2 }2s^{2} 2p^{6}3s^{2} 3p^{3} | 15 | 15 | P |

3d^{6} | 1s^{2 }2s^{2} 2p^{6 }3s^{2 }4s^{2 }3d^{1} | 21 | 21 | SC |

**24. What is the lowest value of n that allows g orbitals to exist?**^{ }

Ans: The lowest value of n that allows g orbitals to exist is 5. This is because the azimuthal quantum number l for g orbitals is 4, and n must be at least l + 1, making n = 5.

**25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron. **

Ans: **For an electron in a 3d orbital, the possible values are:**

n = 3 , l = 2

For l = l,

ml = −2,−1,0, + 1, + 2

**26. An atom of an element contains 29 electrons and 35 neutrons Deduce: **

**(i) The number of protons. and**

Ans: The number of protons in the atom is 29.

**(ii) The electronic configuration of the element. **

Ans: The electronic configuration of the element (At. no. = 29) = 1s^{2 }2s^{2} 2p^{6 }3s^{2 }2p^{6 }3d^{10 }4s^{1.}

**27. Give the number of electrons in the species**.

Ans: H_{2} = _{1}H +_{1}H = 2 electrons

∴ H_{2} + has 2-1 = 1 electrons

O_{2} = _{8}O + _{8}O = 16 electrons

Therefore, O_{2} + has 16- 1 = 15 electrons.

**28. (i) An atomic orbital has n = 3. What are the possible values of l and ml?**

Ans: **The possible values of l and m1 are: **

**(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.**

Ans: for an electron in 2d orbital, the possib;le values of the quantum numbers are = 3,l = 2,

For l = 2, m_{i} = -2, -1, 0, +1, + 2.

**(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.**

Ans: 2s, 2p orbitals are possible.

**29. Using s, p, d notations, describe the orbital with the following quantum numbers.**** **

(a) n = 1, l = 0

(b) n = 3; l = 1

(c) n = 4; l = 2

(d) n = 4; l = 3

Ans:** Using s, p, d notations, the orbitals are:**

(a) n = 1, l = 0: 1s

(b) n = 3, l = 1: 3p

(c) n = 4, l = 2: 4d

(d) n = 4, l = 3: 4f

**30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.**

(a) | n = 0 | l = 0 | m_{l} = 0 | m_{s} = + ½ |

(b) | n = 1 | l = 0 | ml = 0 | ms = + ½ |

(c) | n = 1 | l = 1 | m_{l} = 0 | m_{s }= + ½ |

(d) | n = 2 | l = 1 | m_{l} = 0 | m_{s} = – ½ |

(e) | n = 3 | l = 3 | ml = -3 | ms = – ½ |

(f) | n = 3 | l = 0 | m_{l} = 0 | m_{s }= + ½ |

Ans: (a) The set is not possible. This is because the quantum number n cannot have zero.

(b) This set of quantum numbers is possible.

(c) This set of quantum numbers is possible because when n = 1, l ≠ 1

(d) This set is possible.

(e) Not possible because when n = 3, l ≠ 1.

(f) This set is possible.

**31. How many electrons in an atom may have the following quantum numbers? **

**(a) n = 4, ms = – ½**

Ans: Total electrons in 4th energy level

= 2 × n^{2}

= 2 × 4^{2}

= 32 electrons

Half of them, 16 electrons have m_{s }= – ½.

**(b) n = 3, l = 0.**

Ans: n = 3, l = 0 means 3_{s} orbital which can have electrons.

**32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.**

Ans: Since a hydrogen atom has only one electron, According to Bohr’s principal of quantization of angular momentum (mvr) of a moving electron, we can write,

**When m is the mass of the electron:**

v = The velocity of the electron.

r = radius of the orbit.

h = the planck’s constant.

n = an integral number.

**The de Broglie wavelength λ\lambdaλ of the electron is given by:**

λ= h/p

where p is the momentum of the electron. For an electron

Therefore, the circumference of the orbit is an integral multiple of the de Broglie wavelength.

**33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He + spectrum?**

Ans: **We have to compare wavelength of transition in the H-spectrum with the Balmer transition n = 4 to n = 2 is given:**

∴ For the transition

∴ For hydrogen, spectrum,

**34. Calculate the energy required for the process He**^{+}** (g) He**^{2+}** (g) + e**^{–}**. The ionisation energy for the H atom in the ground state is 2.18 × 10**^{–18}** J atom**^{–1}** **

Ans: Given that,

For H- atom, l.E = E- E_{1}

For the given process, the energy required = E_{n} – E_{1}

= 4 × 2.18 × 10^{-18} J

= 8.72 × 10^{-18} J.

**35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.**

Ans: The diameter of one carbon atom is 0.15 nm.

= 0.15 × 10^{-9} m

= 1.5 × 10^{-10} m

The length of the scale is 20 cm.

= 20 × 10^{-2 }m

= 2 × 10^{-1}m

∴ No. of c-atoms which are to be placed = 20 cm.

= 2 × 10^{-1}m / 1.5 × 10^{-10} m

= 1.33 × 10^{-9.}

**36. 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.**

Ans: Total length = 3.0 cm

Total number of atoms along the length = 2 × 108

**The diameter of carbon atom is:**

= 3.0cm /1 × 108 cm

= 1.5 × 10^{-8} cm

**The radius of carbon atom is:**

= 1.5 × 10^{-8} cm / 2

= 0.75 × 10^{-8 }cm

= 0.75 × 10^{-7 }cm

= 0.075 × 10^{-8 }m

= 0.75 Å 10^{-8 }nm.

**37. The diameter of zinc atom is 2.6 Å. Calculate:**

**(a) radius of zinc atom in pm. and**

Ans: The diameter of a zinc atom is 2.6 Å.

Radius = 2.6 Å/2 = 1.3Å

= 1.3 × 10-10 m

= 130 × 10-12 m

= 130 pm

**Diameter of one atom:**

= 2.6 Å

= 2.6 × 10^{-10} m

**(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.**

Ans: **The number of Zn atoms present on 1.6 cm of length are:**

= 1.6 cm

= 1.6 × 10-2 m

**Calculation of number of atoms:**

= 1.6 × 10-2 m / 2.6 × 10-10

= 6.154 × 10^{7}

**38. A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it.**

Ans: **Change carried by one electron:**

= 1.6022 × 10-19 C

**The number of electrons present in the charged particle:**

(2.5 × 10^{-16}) /(1.6022 × 10^{-19})

= 1560.

**39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10**^{–18}**C, calculate the number of electrons present on it.**

Ans: Charge on the oil drop

= -1.282 × 10^{-18} C

Charge on an electron

= 1.6 × 10 -19 C

∴ Number of electrons in the drop

**40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?**

Ans: In the case of light atoms, the striking a particles will transfer a part of their momentum. The nucleus of heavy elements like gold has a large positive charge and the nucleus of light elements like aluminium have a small positive charge. The repulsion between alpha particles and heavy nuclei is much larger than the repulsion between alpha particles and light nuclei. So if a lighter element like aluminium is used instead of gold, the number of alpha particles deflected will be much less in comparison to the deflections observed when gold was used.

**41. Symbols **_{35}** **^{79}**Br and **^{79}**Br can be written, whereas symbols **_{79}** **^{35 }**Br and **_{35 }**Br are not acceptable. Answer briefly. **

Ans: The symbols _{35} ^{79 }Br and _{35} ^{79 }Br 35Br are acceptable. The former shows the atomic number (35) and mass number (79), while the latter omits the mass number, assuming the atomic number is understood. Symbols like ^{79 }Br and ^{35 }Br are incorrect as they lack essential information.

**42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.**

Ans: Atomic mass number = Number of protons + Number of neutrons

Mass number = 81,i.e. p + n = 81

If protons = x, then neutrons

**43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.**

Ans: Let the number of electrons in the ion carrying a negative charge be x. Then,

Number of neutrons present = x + 111 / 100x = 1.111x

Number of electrons in the neutral atom = (x – 1)

∴ Number of protons in the neutral atom = x – 1

Mass number = number of neutrons + number of protons

∴ 37 = 1.111 + x – 1

Or 2.111x = 38

X = 18

The number of electrons is 18.

The number of protons is 17.

Hence, the symbol of the ion is ^{37}_{17}Cl^{−}.

**44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.**

Ans: Let × be the number of electrons.

The number of protons is x + 3.

The number of neutrons is x + 0.304x = 1.304x.

The mass number is 56.

It is equal to the number of protons and the number of electrons.

56 = x + 3 + 1.304x

2.304 x = 53

x = 23

Therefor no. of protons = Atomic no.

= x + 3 = 23 + 3 = 26

Hence, the symbol of the ion will be ^{56}_{26}Fe^{3+}.

**45. Arrange the following type of radiations in increasing order of frequency:**

**(a) Radiation from microwave oven**.

Ans: Radiation from microwave oven FM radio waves have the lowest frequency among the listed types of radiation.

**(b) amber light from traffic signal.**

Ans: **Amber light from traffic signal:** Microwaves have a higher frequency than FM radio waves but lower than visible light.

**(c) radiation from FM radio**

Ans: **Radiation from FM radio:** Visible light, including amber light, has a higher frequency than microwaves.

**(d) cosmic rays from outer space. and**

Ans: **Cosmic rays from outer space and:** Cosmic rays have the highest frequency among the listed types of radiation.

**(e) X-rays.**

Ans: **X-rays:** X-rays have a much higher frequency compared to visible light.

**Thus, the increasing order of frequency is:** (c) < (a) < (b) < (e) < (d).

**46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser. **

Ans: **The energy E of a photon can be calculated using the formula:**

**47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate:**

**(a) The frequency of emission.**

Ans: Frequency, v = c/λ

c is the speed of light (3.00 × 10^{8} m/s),

λ is the wavelength (616 × 10^{−9} m).

**(b) Distance travelled by this radiation in 30 s.**

Ans: velocity of relationship = 3.0 × 10^{8} ms-1

∴ DIstance travelled in 30s

= 30 × 3 × 10^{8} m

= 9.0 × 10^{9} m

**(c) Energy of quantum. and**

Ans: The energy (E) of a quantum (photon) is given by: E = hν

**where:**

h is Planck’s constant (6.626 × 10^{−34} Js)

νis the frequency we calculated earlier.

**(d) Number of quanta present if it produces 2 J of energy.**

Ans: Number of quanta in 2J energy

**48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10**^{–18}** J from the radiations of 600 nm, calculate the number of photons received by the detector.**

Ans: The energy of 1 photon, E

hv = hc/λ

**where:**

h is Planck’s constant (6.626 × 10^{−34} Js)

c is the speed of light (3.00 × 108 m/)

λ is the wavelength (600 nm or 600 × 10^{−9} m)

Total energy received

= 3.15 × 10^{-8}J

∴ Number of photons received

= 9.51

= 10.

**49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.**

Ans: Frequency, V = 1 / 2 × 10^{-9}s

= 0.5 × 10^{9}s^{-1}

Energy = Nhv = (2.5 × 10^{5)}

(6.626 × 10^{-34} js ) ( 0.5 × 10^{9}s^{-1})

**50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.**

Ans: The frequency of first transition is

The frequency of second transition is

**51. The work function for caesium atom is 1.9 eV. Calculate:**

**(a) the threshold wavelength. and**

Ans: Work function (W_{0}) = h v_{o}

V_{o} = W_{o}/h

**(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron. **

Ans:

K.E (kinetic energy) of ejected electron

**52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate:**

(a) threshold wavelength. and

(b) Planck’s constant.

λ (nm) | 500 | 450 | 400 |

v × 10^{–5} (cm s^{–1}) | 2.55 | 4.35 | 5.20 |

Ans: Suppose threshold wavelength = λ_{o} nm = λ_{o} × 10^{-9} m

**Then:**

**By substituting the results from the three experiments, we obtain:**

Dividing, eq. (ii) (i), we get.

= 2.619 or λ_{o} – 450

= 2.619 λ – 1309.5

Or 1.619 λ_{o} = 859.5

Therefore λ_{o} = 531 nm

Substituting this value in eq. (iii), we get

Or

H = 6.66 × 10^{– 34 }Js.

**53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.**

Ans: Energy of the incident radiation = Work function + Kinetic energy of incident radiation (E)

Hence, kinetic energy of the electron

= 0.35 _{e}v

∴ work function

= 4.83 _{e}v – 0.35 _{e}v

= 4.48 _{e}v

**54. If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus.**

Ans: **he energy of the incident photon is given by:**

E = hc/λ

**The kinetic energy of the ejected electron is given by:**

Energy with which the electron was bound nucleus.

= 7.63 × 10^{3} eV.

**55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 10**^{15}** (Hz) [1/3**^{2}** – 1/n**^{2}**] **

**Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.**

Ans:

= 0.111 – 0.07 = 0.04 = 1/ 25

Or n^{2 }= 25 or n = 5

The radiation with a wavelength of 1285 nm falls within the infrared region.

**56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.**

Ans:

Therefore, if n^{2} = 2 and n^{1} = 5. The transition occurs from the 5th to the 2nd orbit. This transition belongs to the Balmer series.

It lies in the visible region.

**57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 10**^{6}** ms**^{–1}**, calculate de Broglie wavelength associated with this electron.**

Ans: λ = h/mv

**58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.**

Ans: Mass of neutron = 1.675 × 10-27 kg

**59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 10**^{6}** ms**^{–1}**, calculate the de Broglie wavelength associated with it. **

Ans: λ = h/mv

**60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10**^{5}** ms**^{–1}**. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.**

Ans: λ = h/mv

= 3.32 × 10-10m

= 332 pm.

**61. If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/(4π × 0.05) nm, is there any problem in defining this value.**

Ans: Δx = 0.002 nm

= 2 × 10^{-3} nm

= 2 × 10^{-12}m

Δx × Δp = h/4𝝅

Actual momentum

**62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:**

1 | n = 4 | l = 2 | m_{l} = –2 | m_{s} | –1/2 |

2 | n = 3 | l = 2 | m_{l} = 1 | m_{s} | +1/2 |

3 | n = 4 | l = 1 | m_{l} = 0 | m_{s} | +1/2 |

4 | n = 3 | l = 2 | m_{l} = –2 | m_{s} | –1/2 |

5 | n = 3 | l = 1 | m_{l} = –1 | m_{s } | +1/2 |

6 | n = 4 | l = 2 | m_{l} = 0 | m_{s} | +1/2 |

Ans: **The orbitals occupied by the electrons are:**

(i) 4d

(ii) 3d

(iii) 4p

(iv) 3d

(v) 3p

(vi) 4p

**Their energies will be in the order:**

(v) < (ii) = (iv) < (vi) = (iii) < (i).

**63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?**

Ans: The electron in the 4p orbital experiences the lowest effective nuclear charge due to being furthest from the nucleus and experiencing the most shielding from inner electrons.

**64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?**

**(i) 2s and 3s.**

Ans: 2s will experience a larger effective nuclear charge than 3s because it is closer to the nucleus and experiences less shielding.

**(ii) 4d and 4f.**

Ans: 4d will experience a larger effective nuclear charge than 4f due to being less shielded by inner electrons.

**(iii) 3d and 3p.**

Ans: 3p will experience a larger effective nuclear charge than 3d, as 3p is less shielded and closer to the nucleus.

**65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?**

Ans: The unpaired electron in silicon (Si) will experience a greater effective nuclear charge than the unpaired electron in aluminium (Al). This is because silicon has a higher nuclear charge and fewer shielding electrons in the 3p orbital compared to aluminium, resulting in a stronger pull from the nucleus.

**66. Indicate the number of unpaired electrons in:**

**(a) P.**

Ans: _{15}P = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 2p^{1}_{x} 3p^{1}_{y} 3p^{1}_{z}.

No. unpaired electrons = 3

**(b) Si.**

Ans: _{14}Si = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}_{x} 3p^{1}_{y} 3p^{1}_{z}.

No. unpaired electrons = 2

**(c) Cr,**

Ans: _{24}Cr = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{1}.

No. unpaired electrons = 6

**(d) Fe**

Ans: _{26}Fe = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6} 4s^{2}.

No. unpaired electrons= 4.

**(e) Kr.**

Ans:** Krypton (Kr):** Atomic number 36. Electron configuration: All orbitals are filled unpaired electron = 0.

**67. (a) How many subshells are associated with n = 4 ? **

Ans: n = 4 , l = 0,1,2,3, there are four subshells: 4s, 4p, 4d, and 4f.

**(b) How many electrons will be present in the subshells having ms value of –1/2 for n = 4?**

Ans: No. of orbitals in 4th shell = n^{2} = 4^{2} = 16

Each orbital has one electron with m_{s}^{ }= -1/2

Hence, there will be 16 electrons with m_{s}^{ }= -1/2