(ii) Mass of one electron = 9.11×10^-31kg

1 atom of ¹⁴C contains = 14-6 = 8 neutrons.

Nucleus        Z           A         Protons(Z)        Neutrons(A-Z)

¹³C₆                 6           13             6                      13-6=7
¹⁶O₈                 8           16             8                      16-8=8
²⁴Mg₁₂           12          24            12                     24-12=12
⁵⁶Fe₂₆             26         56            26                     56-26=30
 ⁸⁸CSr₃₈          38         88            38                     88-38=50

2.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.

(iii) ⁹Be₄ 

2.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ṽ) of the yellow light.

wave number (ṽ) = 1/λ = 1/580×10^-9 m = 1.72×10⁶ m^-1

2.6. Find energy of each of the photons which
(i) correspond to light of frequency 3×10¹⁵ Hz.
(ii) have wavelength of 0.50 Å.

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

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

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

Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

For the Balmer series, n1 = 2. Hence, ṽ = R(1/2² – 1/n2²)

2.18. 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.

K.E. = 1/2 mv²

∴ v = √2 K.E./m

= 812 ms^-1

By de Broglie equation, λ = h/mv = 6.626×10^-34 Js/(9.11×10^-31 kg) (812 ms^-1) = 8.967×10^-7 m

Isoelectronic are the species having same number of electrons.
A positive charge means the shortage of an electron.

(d) ₉F = 1s²2s²2p⁵

(ii) (a) 3s¹
Completing the electron configuration of the element as 1s²2s²2p⁶3s¹
∴ Number of electrons present in the atom of the element = 2+2+6+1 = 11
∴ Atomic number of the element = 11
(b) 2p³
Completing the electron configuration of the element as 1s²2s²2p³
∴ Number of electrons present in the atom of the element = 2+2+3 = 7
∴ Atomic number of the element = 7
(c) 3p⁵
Completing the electron configuration of the element as 1s²2s²2p⁶3s²3p⁵
∴ Number of electrons present in the atom of the element = 2+2+6+2+5 = 17
∴ Atomic number of the element = 9

For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (m_l) = -2,-1,0,1,2

2.27. Give the number of electrons in the species H₂⁺ , H₂ and O₂⁺

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

Answer

For H-like particles, ṽ = (2π²mZ²e⁴/ch³)×(1/n1² – 1/n2²) = RZ²(1/n1² – 1/n2²)
∴ For He⁺ spectrum, Balmer transition, n=4 to n=2
ṽ = 1/λ = RZ²(1/2² – 1/4²)  = R×4×3/16 = 3R/4
For hydrogen spectrum ,
ṽ = 1/λ = R(1/n1² – 1/n2²) = 3R/4  ⇒ (1/n1² – 1/n2²) = 3/4
which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.

2.34. Calculate the energy required for the process
        He⁺(g) → He²⁺(g) + e^– 
The ionization energy for the H atom in the ground state is 2.18×10^-18 J atom^-1.

Answer

³⁵Br₇₉ is not acceptable because atomic number should be written as subscript, while mass number should be written as superscript. ₃₅Br is not acceptable because atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate mass number.

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

Answer

Mass number = protons+neutrons = p+n = 81 (given)
Let p be x, then neutrons = x + (31.7/100)x = 1.317 x
∴ x + 1.317 x = 81
⇒ 2.317 x = 81
⇒ x = 81/2.317 = 35
Thus, protons = 35 = atomic number.
The symbol of the element is ⁸¹Br₃₅ or ⁸¹₃₅Br

2.43. 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.

Answer

Let the number of electrons in the ion = x
Then, number of neutrons = x+(11.1 x/ 100) = 1.111 x
Number of electrons in the neutral atom = x-1 (ion possesses one unit of negative charge)
∴ Number of protons = x-1
Mass number = No. of protons + No. of neutrons
∴ 1.111 x + x – 1 = 37
⇒ 2.111x = 38
⇒ x = 18
∴ No. of protons = Atomic no. = x-1 = 18-1 = 17
The symbol of the ion is ³⁷₁₇Cl^-1

2.44. 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.

Answer

E = Nhν = Nhc/λ = (5.6×10²⁴)×(6.626×10^-34 Js×3.0×10⁸ ms^-1)/337.1×10^-9 m = 3.3×10⁶ J

2.47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Answer

λ  = 616 nm = 616×10^-9 m
(a) Frequency, ν = c/λ = 3.0×10⁸ ms^-1/616×10^-9 m = 4.87×10¹⁴ s^-1

2.51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer

(a) Work function (W₀) = hν₀
∴ ν₀ = W₀/h = 1.9×1.602×10^-19 J/6.626×10^-34 Js = 4.59×10¹⁴ s^-1            (1eV = 1.602×10^-19 J)
(b) λ₀ = c/ν₀ = 3.0×10⁸ ms^-1/4.59×10¹⁴ s^-1 = 6.54×10^-7 m = 654×10^-9 m = 654 nm
(c) K.E. of ejected electron = h(ν – ν₀) = hc (1/λ – 1/λ₀ )
= (6.626×10^-34 Js×3.0×10⁸ ms^-1)×(1/500×10^-9 m – 1/654×10^-9 m)
= (6.626×3.0×10^-26/10^-9)×(154/500×654) J = 9.36×10^-20 J
K.E. = 1/2 mv² = 9.36×10^-20 J
∴ 1/2×(9.11×10^-31 kg) v² = 9.36×10^-20 kgm²s^-2
⇒ v² = 2.055×10¹¹ m²s^-2 = 20.55×10¹⁰ m²s^-2

(a) Substituting the value of λ and v from the above given data, we get three values of λ₀ as,
λ₀₍₁₎ = 541 nm
λ₀₍₂₎ = 546 nm
λ₀₍₃₎ = 542 nm
Threshold frequency = λ_av = {λ₀₍₁₎+λ₀₍₂₎+λ₀₍₃₎}/3 = (541+546+542)/3 = 543 (approx 540)
(b) Part of this question can’t be solved due to incorrect value of v i.e 5.35.
Students can assume this value as 5.20 if they want to solve this question.

2.53. 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.

Answer

Energy of the incident radiation = Work function + Kinetic energy of photoelectron
E = hc/λ = (6.626×10^-34 Js×3.0×10⁸ ms^-1)/(256.7×10^-9 m) = 7.74×10^-19 J = 4.83 eV
The potential applied gives kinetic energy to the electron.
Hence, kinetic energy of the electron = 0.35 eV
∴ Work Function = 4.83 eV – 0.35 eV = 4.48 eV

2.54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×10⁷ ms^-1 , calculate the energy with which it is bound to the nucleus.

Answer

Energy of the incident photon= hc/λ = (6.626×10^-34 Js×3.0×10⁸ ms^-1)/(150×10^-12m) = 13.25×10^-16 J
Energy of the electron ejected = 1/2 mv² = 1/2×(9.11×10^-31kg)×(1.5×10⁷ms^-1)² = 1.025×10^-16 J
Energy with which the electron was bound to the nucleus = 13.25×10^-16 J – 1.025×10^-16 J
= 12.225×10^-16 J = 12.225×10^-16/1.602×10^-19 eV = 7.63×10³ eV

2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29×10¹⁵(Hz) [1/3²–1/n²] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Answer

ν = c/λ = 3.0×10⁸ ms^-1/1285×10^-9 m = 3.29×10¹⁵ (1/3² – 1/n²)
⇒ 1/n² = 1/9 – (3.0×10⁸ ms^-1/1285×10^-9 m)×(1/3.29×10¹⁵) = 0.111-0.071 = 0.04 = 1/25
⇒ n² = 25
⇒ n = 5
The radiation corresponding to 1285 nm lies in the infrared region.

2.56. 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.

Answer

Radius of nth orbit of H-like particles = 0.529n²/Z Å = 52.9n²/Z pm
r₁ = 1.3225 nm = 1322.5 pm = 52.9n₁²
r₂ = 211.6 pm = 52.9n₂²/Z
∴ r₁/r_{2 =} 1322.5 pm/211.6 pm = n₁²/n₂²
⇒ n₁²/n₂² = 6.25
⇒ n₁/n₂ = 2.5
If n₂ = 2, n₁ = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
ṽ = 1.097×10⁷ m^-1 (1/2² – 1/5²) = 1.097×10⁷×21/100 m^-1
λ = 1/ṽ = 100/(1.097×21×10⁷) m = 434×10^-9 m = 434 nm
It lies in visible range.

2.57. 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 1.6×10⁶ ms^-1 , calculate de Broglie wavelength associated with this electron.

Answer

λ = h/mv = 6.626×10^-34 kgm²s^-1/(9.11×10^-31 kg) (1.6×10⁶ ms^-1) = 4.55×10^-10 m = 455 pm

2.58. 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.

Answer

Mass of neutron = 1.675×10^-27 kg
λ = h/mv
⇒ v = h/mλ = 6.626×10^-34 kgm²s^-1/(1.675×10^-27 kg) (800×10^-12 m) = 4.94×10⁴ ms^-1

2.59. If the velocity of the electron in Bohr’s first orbit is 2.19×10⁶ ms^-1, calculate the de Broglie wavelength associated with it.

2.67. (a) How many sub-shells are associated with n = 4 ? (b) How many electrons will be present in the sub-shells having m_s value of –1/2 for n = 4 ?

Answer

(a) n=4, l = 0, 1, 2, 3. 4 sub-shells are associated with n = 4
(b) No. of orbitals in the shells = n² = 4² = 16
Each orbitals has one electron with m_s = -1/2. Hence, there will be 16 electrons with m_s = -1/2.