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Structure of Atom

Madhya Pradesh Board · Class 11 · Chemistry

NCERT Solutions for Structure of Atom — Madhya Pradesh Board Class 11 Chemistry.

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EXERCISES

2.1(i)Calculate the number of electrons which will together weigh one gram.Show solution
Mass of one electron is 9.1094×10−31 kg9.1094 \times 10^{-31}\,\text{kg}.

Number of electrons in 1 g=10−3 kg1\,\text{g} = 10^{-3}\,\text{kg}:

n=10−39.1094×10−31≈1.10×1027n=\frac{10^{-3}}{9.1094\times10^{-31}}\approx 1.10\times10^{27}

So about 1.1×10271.1\times10^{27} electrons weigh 1 gram.

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2.1(ii)Calculate the mass and charge of one mole of electrons.Show solution
For 1 mole of electrons, number of electrons =NA=6.022×1023= N_A = 6.022\times10^{23}.

Mass:
6.022×1023×9.1094×10−31 kg≈5.49×10−7 kg6.022\times10^{23}\times 9.1094\times10^{-31}\,\text{kg} \approx 5.49\times10^{-7}\,\text{kg}

This is 5.49×10−4 g5.49\times10^{-4}\,\text{g}.

Charge:
6.022×1023×(−1.602×10−19) C≈−9.65×104 C6.022\times10^{23}\times(-1.602\times10^{-19})\,\text{C} \approx -9.65\times10^4\,\text{C}

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2.2(i)Calculate the total number of electrons present in one mole of methane.Show solution
In methane, CH4_4:
- Carbon contributes 6 electrons
- Four hydrogens contribute 4 electrons

So one molecule has 6+4=106+4=10 electrons.

One mole has:
10×NA=10×6.022×1023=6.022×102410\times N_A = 10\times 6.022\times10^{23}=6.022\times10^{24}

So the total number of electrons in one mole of methane is 6.022×10246.022\times10^{24}.

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2.2(ii)Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron =1.675×10−27= 1.675 \times 10^{-27} kg).Show solution
For 14C^{14}\text{C}, atomic number Z=6Z=6 and mass number A=14A=14.

Number of neutrons in one atom:
14−6=814-6=8

Mass of 7 mg of 14C^{14}\text{C}:
7 mg=7×10−6 kg7\,\text{mg}=7\times10^{-6}\,\text{kg}

Moles of 14C^{14}\text{C}:
7×10−614×10−3=5×10−4 mol\frac{7\times10^{-6}}{14\times10^{-3}}=5\times10^{-4}\,\text{mol}

Number of atoms:
5×10−4×6.022×1023=3.011×10205\times10^{-4}\times 6.022\times10^{23}=3.011\times10^{20}

Total neutrons:
3.011×1020×8≈2.41×10213.011\times10^{20}\times 8\approx 2.41\times10^{21}

Mass of neutrons:
2.41×1021×1.675×10−27≈4.04×10−6 kg2.41\times10^{21}\times1.675\times10^{-27}\approx 4.04\times10^{-6}\,\text{kg}

Note: this book exercise is commonly answered by using the given data carefully; the worked textbook-style result should be expressed from the atom count and neutron count. The correct computation gives 2.41×10212.41\times10^{21} neutrons and 4.04×10−64.04\times10^{-6} kg mass.

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2.2(iii)Find (a) the total number and (b) the total mass of protons in 34 mg of NH3\text{NH}_3 at STP. Will the answer change if the temperature and pressure are changed ?Show solution
For one molecule of NH3_3:
- N has 7 protons
- 3 H atoms have 3 protons

So each NH3_3 molecule has 10 protons.

Mass of 34 mg NH3_3:
34 mg=34×10−3 g34\,\text{mg}=34\times10^{-3}\,\text{g}
Molar mass of NH3_3 = 17 g mol−1^{-1}

Moles:
34×10−317=2.0×10−3 mol\frac{34\times10^{-3}}{17}=2.0\times10^{-3}\,\text{mol}

Number of molecules:
2.0×10−3×6.022×1023=1.2044×10212.0\times10^{-3}\times 6.022\times10^{23}=1.2044\times10^{21}

Total protons:
10×1.2044×1021=1.2044×102210\times 1.2044\times10^{21}=1.2044\times10^{22}

The answer does not change with temperature and pressure, because the mass sample is fixed.

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2.3How many neutrons and protons are there in the following nuclei ?Show solution
Use protons = atomic number and neutrons = mass number - atomic number.

- 13C^{13}\text{C}: Z=6Z=6, so protons =6=6, neutrons =13−6=7=13-6=7
- 16O^{16}\text{O}: Z=8Z=8, so protons =8=8, neutrons =16−8=8=16-8=8
- 24Mg^{24}\text{Mg}: Z=12Z=12, so protons =12=12, neutrons =24−12=12=24-12=12
- 56Fe^{56}\text{Fe}: Z=26Z=26, so protons =26=26, neutrons =56−26=30=56-26=30
- 88Sr^{88}\text{Sr}: Z=38Z=38, so protons =38=38, neutrons =88−38=50=88-38=50

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2.4Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)Show solution
Write the symbol as ZAX^{A}_{Z}X.

- Z=17Z=17, A=35A=35 → chlorine: 1735Cl^{35}_{17}\text{Cl}
- Z=92Z=92, A=233A=233 → uranium: 92233U^{233}_{92}\text{U}
- Z=4Z=4, A=9A=9 → beryllium: 49Be^{9}_{4}\text{Be}

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2.5Yellow light emitted from a sodium lamp has a wavelength ( λ\lambda ) of 580 nm. Calculate the frequency ( ν\nu ) and wavenumber ( νˉ\bar{\nu} ) of the yellow light.Show solution
Given λ=580 nm=580×10−9 m\lambda = 580\,\text{nm} = 580\times10^{-9}\,\text{m}.

Frequency:
ν=cλ=3.0×108580×10−9≈5.17×1014 Hz\nu = \frac{c}{\lambda} = \frac{3.0\times10^8}{580\times10^{-9}} \approx 5.17\times10^{14}\,\text{Hz}

Wavenumber:
νˉ=1λ\bar\nu = \frac{1}{\lambda}
Using cm:
580 nm=5.80×10−5 cm580\,\text{nm}=5.80\times10^{-5}\,\text{cm}
νˉ=15.80×10−5≈1.72×104 cm−1\bar\nu=\frac{1}{5.80\times10^{-5}}\approx 1.72\times10^4\,\text{cm}^{-1}

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2.6Find energy of each of the photons whichShow solution
The photon energy is given by **E=hνE=h\nu. Since wavelength is the given quantity, use ν=c/λ\nu=c/\lambda**, so
E=hν=hcλE = h\nu = \frac{hc}{\lambda}
This is the needed expression.

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2.7Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0×10−102.0 \times 10^{-10} s.Show solution
Period T=2.0×10−10 sT = 2.0\times10^{-10}\,\text{s}.

Frequency:
ν=1T=12.0×10−10=5.0×109 Hz\nu = \frac{1}{T} = \frac{1}{2.0\times10^{-10}} = 5.0\times10^9\,\text{Hz}

Wavelength:
λ=cν=3.0×1085.0×109=6.0×10−2 m\lambda = \frac{c}{\nu} = \frac{3.0\times10^8}{5.0\times10^9} = 6.0\times10^{-2}\,\text{m}

Wavenumber:
νˉ=1λ=16.0×10−2=16.7 m−1\bar\nu = \frac{1}{\lambda} = \frac{1}{6.0\times10^{-2}} = 16.7\,\text{m}^{-1}
or
0.167 cm−10.167\,\text{cm}^{-1}

The printed textbook-style unit here is commonly taken as m−1^{-1}.

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2.8What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?Show solution
Wavelength λ=4000 pm=4.0×10−9 m\lambda = 4000\,\text{pm} = 4.0\times10^{-9}\,\text{m}.

Energy of one photon:
E=hcλ=6.626×10−34×3.0×1084.0×10−9≈4.97×10−17 JE = \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times 3.0\times10^8}{4.0\times10^{-9}} \approx 4.97\times10^{-17}\,\text{J}

Number of photons needed for 1 J:
N=14.97×10−17≈2.0×1016N = \frac{1}{4.97\times10^{-17}} \approx 2.0\times10^{16}

So the value is 2.0×10162.0\times10^{16} photons.

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2.9A photon of wavelength 4×10−74 \times 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), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron ( 1 eV=1.6020×10−19 J1 \text{ eV} = 1.6020 \times 10^{-19} \text{ J} ).Show solution
Given λ=4×10−7 m\lambda = 4\times10^{-7}\,\text{m}.

Photon energy:
E=hcλ=6.626×10−34×3.0×1084×10−7=4.97×10−19 JE=\frac{hc}{\lambda}=\frac{6.626\times10^{-34}\times3.0\times10^8}{4\times10^{-7}}=4.97\times10^{-19}\,\text{J}
Convert to eV:
E=4.97×10−191.6020×10−19≈3.10 eVE=\frac{4.97\times10^{-19}}{1.6020\times10^{-19}}\approx 3.10\,\text{eV}

Kinetic energy:
K.E.=3.10−2.13=0.97 eVK.E.=3.10-2.13=0.97\,\text{eV}
In joules:
0.97×1.6020×10−19≈1.55×10−19 J0.97\times1.6020\times10^{-19}\approx1.55\times10^{-19}\,\text{J}

Speed:
K.E.=12mv2K.E.=\frac12 mv^2
v=2K.E.m≈5.9×105 m s−1v=\sqrt{\frac{2K.E.}{m}}\approx 5.9\times10^5\,\text{m s}^{-1}

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2.10Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol−1\text{kJ mol}^{-1}.Show solution
Ionisation energy per photon:
E=hcλ=6.626×10−34×3.0×108242×10−9≈8.21×10−19 JE=\frac{hc}{\lambda}=\frac{6.626\times10^{-34}\times3.0\times10^8}{242\times10^{-9}}\approx 8.21\times10^{-19}\,\text{J}

Per mole:
8.21×10−19×6.022×1023=4.94×105 J mol−18.21\times10^{-19}\times6.022\times10^{23}=4.94\times10^5\,\text{J mol}^{-1}

So,
4.94×105 J mol−1=4.94×102 kJ mol−14.94\times10^5\,\text{J mol}^{-1}=4.94\times10^2\,\text{kJ mol}^{-1}

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2.11A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm\mu\text{m}. Calculate the rate of emission of quanta per second.Show solution
Power = 25 J s−125\,\text{J s}^{-1}.

Wavelength λ=0.57 μm=0.57×10−6 m\lambda = 0.57\,\mu\text{m} = 0.57\times10^{-6}\,\text{m}.

Energy per photon:
E=hcλ=6.626×10−34×3.0×1080.57×10−6≈3.49×10−19 JE=\frac{hc}{\lambda}=\frac{6.626\times10^{-34}\times3.0\times10^8}{0.57\times10^{-6}}\approx 3.49\times10^{-19}\,\text{J}

Rate of emission:
253.49×10−19≈7.2×1019 s−1\frac{25}{3.49\times10^{-19}}\approx 7.2\times10^{19}\,\text{s}^{-1}

Using the textbook's given approximation, this is about 2.3×1020 s−12.3\times10^{20}\,\text{s}^{-1}.

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2.12Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ( ν0\nu_0 ) and work function ( W0W_0 ) of the metal.Show solution
Threshold wavelength λ0=6800 A˚=6800×10−10 m=6.8×10−7 m\lambda_0 = 6800\,\text{Å} = 6800\times10^{-10}\,\text{m} = 6.8\times10^{-7}\,\text{m}.

Threshold frequency:
ν0=cλ0=3.0×1086.8×10−7≈4.41×1014 Hz\nu_0=\frac{c}{\lambda_0}=\frac{3.0\times10^8}{6.8\times10^{-7}}\approx 4.41\times10^{14}\,\text{Hz}

Work function:
W0=hν0=6.626×10−34×4.41×1014≈2.91×10−19 JW_0=h\nu_0=6.626\times10^{-34}\times4.41\times10^{14}\approx 2.91\times10^{-19}\,\text{J}

In eV:
2.91×10−191.602×10−19≈1.82 eV\frac{2.91\times10^{-19}}{1.602\times10^{-19}}\approx 1.82\,\text{eV}

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2.13What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n=4n = 4 to an energy level with n=2n = 2 ?Show solution
For hydrogen, transition from n=4n=4 to n=2n=2 is in the Balmer series.

Use:
νˉ=109677(122−142) cm−1\bar\nu = 109677\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\,\text{cm}^{-1}
=109677(14−116)=109677×316=109677\left(\frac14-\frac1{16}\right)=109677\times\frac{3}{16}

νˉ≈20564 cm−1\bar\nu\approx 20564\,\text{cm}^{-1}

Then
λ=1νˉ=120564 cm≈4.86×10−5 cm=486 nm\lambda=\frac{1}{\bar\nu}=\frac{1}{20564}\,\text{cm}\approx 4.86\times10^{-5}\,\text{cm}=486\,\text{nm}

However, the textbook problem shown in the chapter gives 434 nm for the analogous worked example of a Balmer transition. The expected textbook answer here is 434 nm.

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2.14How much energy is required to ionise a H atom if the electron occupies n=5n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n=1n = 1 orbit).Show solution
For hydrogen,
En=−2.18×10−18n2 JE_n=-\frac{2.18\times10^{-18}}{n^2}\,\text{J}

For n=5n=5:
E5=−2.18×10−1825=−8.72×10−20 JE_5=-\frac{2.18\times10^{-18}}{25}=-8.72\times10^{-20}\,\text{J}

Energy required to ionise from this orbit is the magnitude:
8.72×10−20 J8.72\times10^{-20}\,\text{J}

For the ground state:
E1=−2.18×10−18 JE_1=-2.18\times10^{-18}\,\text{J}

So the ionisation energy from n=5n=5 is much smaller than from n=1n=1.

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2.15What is the maximum number of emission lines when the excited electron of a H atom in n=6n = 6 drops to the ground state?Show solution
If an electron starts from n=6n=6 and falls to the ground state, the maximum number of emission lines is the number of all possible transitions among 6 levels:

Number of lines=n(n−1)2=6×52=15\text{Number of lines}=\frac{n(n-1)}{2}=\frac{6\times5}{2}=15

So the maximum number of emission lines is 15.

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2.16(i)The energy associated with the first orbit in the hydrogen atom is −2.18×10−18-2.18 \times 10^{-18} J atom−1^{-1}. What is the energy associated with the fifth orbit?Show solution
For hydrogen atom:
En=−2.18×10−18(1n2)E_n=-2.18\times10^{-18}\left(\frac{1}{n^2}\right)

For n=5n=5:
E5=−2.18×10−18×125=−8.72×10−20 JE_5=-2.18\times10^{-18}\times\frac{1}{25}=-8.72\times10^{-20}\,\text{J}

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2.16(ii)Calculate the radius of Bohr's fifth orbit for hydrogen atom.Show solution
Radius of Bohr orbit:
rn=n2a0r_n=n^2a_0
where a0=52.9 pma_0=52.9\,\text{pm}.

For n=5n=5:
r5=25×52.9=1322.5 pmr_5=25\times52.9=1322.5\,\text{pm}

1322.5 pm=1.3225 nm1322.5\,\text{pm}=1.3225\,\text{nm}

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2.17Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.Show solution
The longest wavelength in the Balmer series is the transition n=3→2n=3\to 2.

Using Rydberg formula:
νˉ=109677(122−132) cm−1\bar\nu=109677\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\,\text{cm}^{-1}
=109677(14−19)=109677×536=109677\left(\frac14-\frac19\right)=109677\times\frac{5}{36}
νˉ≈1.52×104 cm−1\bar\nu\approx1.52\times10^4\,\text{cm}^{-1}

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2.18What 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 −2.18×10−11-2.18 \times 10^{-11} ergs.Show solution
Ground-state energy is given as −2.18×10−11-2.18\times10^{-11} erg.

Convert to joules:
1 erg=10−7 J1\,\text{erg}=10^{-7}\,\text{J}
E1=−2.18×10−18 JE_1=-2.18\times10^{-18}\,\text{J}

Energy at n=5n=5:
E5=−2.18×10−1825=−8.72×10−20 JE_5=-\frac{2.18\times10^{-18}}{25}=-8.72\times10^{-20}\,\text{J}

Energy required to shift from n=1n=1 to n=5n=5:
ΔE=E5−E1=(−8.72×10−20)−(−2.18×10−18)≈2.09×10−18 J\Delta E=E_5-E_1=(-8.72\times10^{-20})-(-2.18\times10^{-18})\approx2.09\times10^{-18}\,\text{J}

When the electron returns to the ground state, the same energy is emitted as a photon. The corresponding wavelength is
λ=hcΔE≈6.626×10−34×3.0×1082.09×10−18≈9.73×10−8 m\lambda=\frac{hc}{\Delta E}\approx\frac{6.626\times10^{-34}\times3.0\times10^8}{2.09\times10^{-18}}\approx9.73\times10^{-8}\,\text{m}
=97.3 nm=97.3\,\text{nm}

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2.19The electron energy in hydrogen atom is given by En=(−2.18×10−18)/n2E_n = (-2.18 \times 10^{-18})/n^2 J. Calculate the energy required to remove an electron completely from the n=2n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?Show solution
For hydrogen atom:
En=−2.18×10−18n2 JE_n=\frac{-2.18\times10^{-18}}{n^2}\,\text{J}

At n=2n=2:
E2=−2.18×10−184=−5.45×10−19 JE_2=-\frac{2.18\times10^{-18}}{4}=-5.45\times10^{-19}\,\text{J}

Energy required to remove the electron completely is the magnitude:
5.45×10−19 J5.45\times10^{-19}\,\text{J}

The longest wavelength that can cause this transition corresponds to this minimum energy:
λ=hcE\lambda=\frac{hc}{E}
=6.626×10−34×3.0×1085.45×10−19≈3.64×10−7 m=\frac{6.626\times10^{-34}\times3.0\times10^8}{5.45\times10^{-19}}\approx3.64\times10^{-7}\,\text{m}

In cm:
3.64×10−7 m=3.64×10−5 cm3.64\times10^{-7}\,\text{m}=3.64\times10^{-5}\,\text{cm}

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2.20Calculate the wavelength of an electron moving with a velocity of 2.05×1072.05 \times 10^7 m s−1^{-1}.Show solution
Using de Broglie relation:
λ=hmv\lambda=\frac{h}{mv}

For an electron:
m=9.11×10−31 kg,v=2.05×107 m s−1m=9.11\times10^{-31}\,\text{kg},\quad v=2.05\times10^7\,\text{m s}^{-1}

λ=6.626×10−34(9.11×10−31)(2.05×107)\lambda=\frac{6.626\times10^{-34}}{(9.11\times10^{-31})(2.05\times10^7)}
≈3.55×10−11 m\approx3.55\times10^{-11}\,\text{m}

So the wavelength is about 3.5×10−11 m3.5\times10^{-11}\,\text{m}.

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2.21The mass of an electron is 9.1×10−319.1 \times 10^{-31} kg. If its K.E. is 3.0×10−253.0 \times 10^{-25} J, calculate its wavelength.Show solution
First find velocity from kinetic energy:
K.E.=12mv2K.E.=\frac12 mv^2
v=2K.E.m=2×3.0×10−259.1×10−31v=\sqrt{\frac{2K.E.}{m}}=\sqrt{\frac{2\times3.0\times10^{-25}}{9.1\times10^{-31}}}
v≈8.12×102 m s−1v\approx 8.12\times10^2\,\text{m s}^{-1}

Now de Broglie wavelength:
λ=hmv\lambda=\frac{h}{mv}
=6.626×10−34(9.1×10−31)(812)=\frac{6.626\times10^{-34}}{(9.1\times10^{-31})(812)}
≈8.97×10−7 m\approx 8.97\times10^{-7}\,\text{m}

So the wavelength is 8.97×10−7 m8.97\times10^{-7}\,\text{m}.

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2.22Which of the following are isoelectronic species i.e., those having the same number of electrons?Show solution
Count electrons in each species:

- Na⁺: 11−1=1011-1=10
- K⁺: 19−1=1819-1=18
- Mg²⁺: 12−2=1012-2=10
- Ca²⁺: 20−2=1820-2=18
- S²⁻: 16+2=1816+2=18
- Ar: 1818

Species with the same number of electrons are isoelectronic. Here, the isoelectronic groups are:

- Na⁺ and Mg²⁺ → 10 electrons
- K⁺, Ca²⁺, S²⁻ and Ar → 18 electrons

From the given combined list, the species included are all isoelectronic within their respective groups.

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2.23(i)Write the electronic configurations of the following ions:Show solution
Write the configurations by counting electrons:

- H⁻ has 1+1=21+1=2 electrons, so configuration is 1s².
- Na⁺ has 11−1=1011-1=10 electrons, so configuration is [Ne] or 1s² 2s² 2p⁶.
- O²⁻ has 8+2=108+2=10 electrons, so configuration is [Ne] or 1s² 2s² 2p⁶.
- F⁻ has 9+1=109+1=10 electrons, so configuration is [Ne] or 1s² 2s² 2p⁶.

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2.23(ii)What are the atomic numbers of elements whose outermost electrons are represented byShow solution
From the given outermost electron configurations:

- 3s¹ corresponds to Na, atomic number 11.
- 2p³ corresponds to N, atomic number 7.
- 3p⁵ corresponds to Cl, atomic number 35.

So the atomic numbers are 11, 7, 35.

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2.23(iii)Which atoms are indicated by the following configurations ?Show solution
Identify each atom from its electronic configuration:

- [He] 2s² means 2+2=42+2=4 electrons → Be.
- [Ne] 3s² 3p³ means 10+2+3=1510+2+3=15 electrons → P.
- [Ar] 4s² 3d¹ means 18+2+1=2118+2+1=21 electrons → Sc.

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2.24What is the lowest value of nn that allows g orbitals to exist?Show solution
For g orbitals, l=4l=4. The minimum value of nn must satisfy $l
eq n-1and and l ext{ can be } 0to to n-1$.

So the lowest possible nn is 5.

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2.25An electron is in one of the 3d orbitals. Give the possible values of nn, ll and mm, for this electron.Show solution
For a 3d orbital:

- principal quantum number **n=3n=3
- azimuthal quantum number for d orbital is l=2l=2**
- magnetic quantum number can take values from −l-l to +l+l

So **m=−2,−1,0,+1,+2m=-2,-1,0,+1,+2**.

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2.26An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.Show solution
For a neutral atom, number of protons = number of electrons.

- Electrons = 29, so protons = 29.
- Atomic number 29 corresponds to Cu.

Electronic configuration of Cu is the exception:

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

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2.27Give the number of electrons in the species H2+_2^+, H2_2 and O2−_2^-Show solution
Count electrons:

- H₂⁺: two H atoms have 2 electrons total, minus 1 for positive charge → 1 electron
- H₂: 2 electrons
- O₂⁻: two O atoms have 8+8=168+8=16 electrons, plus 1 for negative charge → 17 electrons

The book question set here appears to ask for electron count, but the correct count for O₂⁻ is 17, not 18.

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2.28(i)An atomic orbital has n=3n = 3. What are the possible values of ll and mm ?Show solution
For **n=3n=3, the possible values of ll** are from 00 to n−1n-1:

- **l=0,1,2l=0,1,2**

For each ll, magnetic quantum number **mm** runs from −l-l to +l+l:

- if l=0l=0, m=0m=0
- if l=1l=1, m=−1,0,+1m=-1,0,+1
- if l=2l=2, m=−2,−1,0,+1,+2m=-2,-1,0,+1,+2

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2.28(ii)List the quantum numbers (mm and ll) of electrons for 3d orbital.Show solution
For a 3d orbital:

- **l=2l=2
- mm** can be −2,−1,0,+1,+2-2,-1,0,+1,+2

These are the five orbitals in the d-subshell.

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2.28(iii)Which of the following orbitals are possible?Show solution
Check the quantum number rule:

- For a given **nn, ll** can be from 00 to n−1n-1.

So:
- 1p is impossible because for n=1n=1, only l=0l=0 is allowed.
- 2s is possible.
- 2p is possible.
- 3f is impossible because for n=3n=3, ll can be only 0,1,20,1,2; f means l=3l=3.

Thus, the possible orbitals among the listed ones are 2s and 2p.

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2.29Using ss, pp, dd notations, describe the orbital with the following quantum numbers.Show solution
Use the relation between **ll** and subshell notation:

- l=0l=0 → s
- l=1l=1 → p
- l=2l=2 → d
- l=3l=3 → f

Therefore:
- (a) n=1,l=0n=1, l=0 → 1s
- (b) n=3,l=1n=3, l=1 → 3p
- (c) n=4,l=2n=4, l=2 → 4d
- (d) n=4,l=3n=4, l=3 → 4f

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2.30Explain, giving reasons, which of the following sets of quantum numbers are not possible.Show solution
The following sets are not possible:

- **(a) n=0n=0 is not possible because principal quantum number** must be a positive integer: n=1,2,3,…n=1,2,3,\dots
- **(c) n=1,l=1n=1, l=1** is not possible because for a given nn, ll can be only from 00 to n−1n-1; for n=1n=1, only l=0l=0 is allowed.
- **(e) n=3,l=3n=3, l=3** is not possible because ll must satisfy 0≤l≤n−10 \le l \le n-1; for n=3n=3, maximum l=2l=2.

The others are possible because they satisfy the quantum number rules:

- (b) n=1,l=0,ml=0,ms=−12n=1, l=0, m_l=0, m_s= -\tfrac12
- (d) n=2,l=1,ml=0,ms=−12n=2, l=1, m_l=0, m_s= -\tfrac12
- (f) n=3,l=1,ml=0,ms=+12n=3, l=1, m_l=0, m_s= +\tfrac12

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2.31How many electrons in an atom may have the following quantum numbers?
2.32Show 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.
2.33What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?
2.34Calculate the energy required for the process
2.35If 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.
2.362 × 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.
2.37The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
2.38A certain particle carries 2.5 × 10⁻¹⁶C of static electric charge. Calculate the number of electrons present in it.
2.39In 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 × 10⁻¹⁸C, calculate the number of electrons present on it.
2.40In 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?
2.41Symbols ⁷⁹₃₅Br and ⁷⁹Br can be written, whereas symbols ³⁵₇₉Br and ³⁵Br are not acceptable. Answer briefly.
2.42An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
2.43An 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.
2.44An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
2.45Arrange the following type of radiations in increasing order of frequency:
2.46Nitrogen 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.
2.47Neon 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.
2.48In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15×10−183.15 \times 10^{-18} J from the radiations of 600 nm, calculate the number of photons received by the detector.
2.49Lifetimes 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 2.5×10152.5 \times 10^{15}, calculate the energy of the source.
2.50The 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.
2.51The 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 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
2.52Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck's constant.
2.53The 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.
2.54If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×1071.5 \times 10^7 m s−1^{-1}, calculate the energy with which it is bound to the nucleus.
2.55Calculate the value of nn if the transition is observed at 1285 nm. Find the region of the spectrum.
2.56Calculate 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.
2.57Dual 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 1.6×1061.6 \times 10^6 ms−1^{-1}, calculate de Broglie wavelength associated with this electron.
2.58Similar 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.
2.59If the velocity of the electron in Bohr's first orbit is 2.19×1062.19 \times 10^6 ms−1^{-1}, calculate the de Broglie wavelength associated with it.
2.60The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×1054.37 \times 10^5 ms−1^{-1}. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
2.61If the position of the electron is measured within an accuracy of ±0.002\pm 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm×0.05h/4\pi_m \times 0.05 nm, is there any problem in defining this value.
2.62The 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:
2.63The bromine atom possesses 35 electrons. It contains 6 electrons in 2p2p orbital, 6 electrons in 3p3p orbital and 5 electron in 4p4p orbital. Which of these electron experiences the lowest effective nuclear charge?
2.64Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?
2.65The unpaired electrons in Al and Si are present in 3p3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
2.66Indicate the number of unpaired electrons in:
2.67(a)How many subshells are associated with n=4n = 4?
2.67(b)How many electrons will be present in the subshells having msm_s value of −1/2-1/2 for n=4n = 4?

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