**1) An electron from certain higher orbit (x) upon de-excitation to first excited state of He ^{+} forms a maximum of 6 spectral lines. The wavelength associated with the electron in the above higher orbit x is observed to be:**

(a) **8.31 Å ** (b) **11.63 Å** (c) **6.65 Å** (d) **13.3 Å**

(d) Maximum number of spectral lines if an electron jumps from ‘n_{2}’ level to ‘n_{1}’ level in a mono-electronic species = n_{2} – n_{1}

x – 2 = 6 ⇒ x = 8

From de Broglie hypothesis and Bohr’s theory,

2πr = nλ [n no.oforbit]

2πr = xλ

λ = 2πr/x

λ = (2×3.14×0.529×64)/(8×2)

λ = 13.3Å

Alternatively, de Broglie wavelength (λ) can also be calculated by λ = h/mv

Where , (2.18×10^{6})/n m/s

λ = h/mv = (6.629×10^{-34}×8)/(9.1×10^{-31}×2.18×10^{6}×2)

λ = 13.3Å

**2) Consider the following statements: **

**I. If ionization energy of He ^{+} is 19.6×10^{-18} J per atom, then the energy of second stationary state in Li^{2+} ion is – 11.025×10^{-18}J. **

**II. The difference in energy between the Ist and IInd Bohr orbit of hydrogen atom is 1.635 × 10 ^{-8} J (RH = 109677 cm^{-1}, h = 6.6 × 10^{-34}Js, C = 3 × 10^{8}ms^{-1}) **

**III. According to Bohr’s Theory, Ionization energy of Li ^{+} ion is 1180 KJ mol^{-1}. **

**IV. The wavelength number for the shortest wavelength transition in Balmer series of atomic hydrogen is 27758 cm ^{-1}. **

**Now, identify the correct statements from the above: **

(a) **I, II, III, IV ** (b) **I, II** (c) **III & IV** (d) **I, II, IV **

(b) I. Energy of first Bohr orbit in one electron species given as:

E = Z^{2} × E_{H}

For He^{+}, Z = 2

∴ E_{H} = E/Z^{2} = -(19.6×10^{-18})/4 J

Now, for Li^{2+} (Z = 3)

E = (3)^{2} × -(19.6×10^{-18})/4 = -44.1 × 10^{-18} J

∴ for the 2^{nd} Bohr orbit,

E_{2} = E_{1}/n^{2} = (-44.1×10^{-18})/4 = 11.025×10^{-18} J

II. ΔE = R_{H}hcZ^{2 }(1/n_{1}^{2} – 1/n_{2}^{2})

= 1.09672×10^{7}×6.6×10^{-34}×3×10^{8}(1/1^{2} – 1/2^{2})

= 1.635 × 10^{-8} J

III. Bohr’s theory in not true for Li^{+} as it contains two electrons.

IV. 1/λ = R_{H}Z^{2} (1/n_{1}^{2} – 1/n_{2}^{2})

For shortest wavelength in Balmer series, n_{1} = 2, n_{2} = ∞

∴ 1/λ = 1.09677×1^{2}(1/2^{2} – 1/∞^{2}) = 27419.25 cm^{-1}

**3) In Bohr series of lines of hydrogen spectrum, the third line from red end corresponds to which one of the following inter-orbit jumps of the electrons for Bohr orbits in an atom of hydrogen **

(a) **3 → 2 ** (b) **5 → 2** (c) **4 → 1** (d) **2 → 5**

(b)

**4) When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de Broglie wavelength, λ _{A} . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T_{B}(T_{B} = T_{A} – 1.50eV) . If the de Broglie wavelength of these photoelectrons is λ_{B}(λ_{B} = 2λ_{A}), then which is not correct?**

(a) **The work function of A is 2.25 eV** ** ** (b) **the work function of B is 3.70 eV**

(c) **T _{A} = 2.00 eV** (d)

**T**

_{B}= 2.75 eV(d) λ = h/mu ⇒ u = h/λm

u_{A} = (h /λ_{A})m

u_{B} = (h/λ_{B})m

∴ u_{A}/u_{B }= λ_{B}/λ_{A }= 2 (λ_{B }= 2λ_{A})

Now, T_{A}/T_{B} = mu_{A}^{2}/mu_{B}^{2} = 4/1

T_{A }= 4T_{B}

Also T_{A }– T_{B} = 1.50

4T_{B }– T_{B} = 1.50

T_{B} = 0.50

T_{A} = 0.50 + 1.50 = 2.00 eV

Also, 4.25 = Φ_{A}+T_{A}

∴ Φ_{A }= 4.24 – 2.00 = 2.25eV

4.20 = Φ_{B}+T_{B}

Φ_{B }= 4.20 – 0.50 = 3.70 eV

**5) Consider one He ^{+} ion in excited state (n = 5). Which of the following observations will hold true as per the Bohr’s model? **

(a) **10 emission spectral lines will be seen **

(b) **The ionization energy needed is less than 2 eV**

(c) **The longest emitted wavelength is less than 10/R (R = Rydberg’s constant)**

(d) **The electronic separation from the centre of nucleus is more than 6Å **

(d)

(a) A multitude of He^{+} will produce 10 possible emission spectral lines. However, a single He^{+} can produce maximum 4 lines (not more)

5 → 4 → 3 → 2 → 1

(b) E_{n} = -13.6(Z^{2}/n^{2}) = -13.6(2^{2}/5^{2}) = -13.6(4/25) = -2.176 eV

1E = + 2.176 eV > 2 eV

1/λ = RZ^{2} (1/n_{1}^{2} – 1/n_{2}^{2})

(c) 1/λ = R4 (1/4^{2} – 1/5^{2})

1/λ = 4R [9/(16×25)]

λ = (16×25)/(4×9×R) = 100/9R > 10/R

(d) r_{n} = 0.53 (n^{2}/Z) = 0.53 (5^{2}/Z) = 6.625Å > 6Å

**6) Which of the following sets of quantum numbers represents the highest energy of an atom? **

(a) **n = 4, l = 0, m = 0, s = +1/2 **(b) **n = 3, l = 0, m = 9, s = +1/2**

(c) **n = 3, l = 1, m = 1, s = +1/2 **(d) **n = 3, l = 2, m = 1, s = +1/2**

(d) Energy of an electron is given by ‘n+l’ rule which is “higher the value of ‘n+l’, higher the energy of electron and for same value of ‘n+l’, higher to ‘n’ value higher the energy of electron”.

(a) ‘n+l’ = 4 + 0 = 4 (4s)

(b) ‘n+l’ = 3 + 0 = 3 (3s)

(c) ‘n+l’ = 3 + 1 = 4 (3p)

(d) ‘n+l’ = 3 + 2 = 5 (3d)

Hence 3d has the highest energy amongst others.

**7) A 1 kW radio transmitter operates at a frequency of 880 Hz. How many quanta per second does it emit? **

(a) **1.71 × 10 ^{21} ** (b)

**1.71 × 10**(c)

^{30}**6.02 × 10**(d)

^{23}**2.85 × 10**

^{26 }(b) A 1kW radio transmitter emits 1000 Joules of radio waves per second.

Energy of one quanta, E = hν

= 6.626×10^{-34}×880×10^{3} = 5.83×10^{-28} J

Number of quanta emitted per second = Energy emitted per sec/Energy of one quanta

= 1000(5.83×10^{-28}) = 1.71×10^{30}

**8) The orbital diagram in which both the Pauli’s exclusion principle and Hund’s rule are violated is **

(a) **↑↓ ↑↑,↑ ** (b) **↑↓ ↑↓, ,↑↓** (c) **↑↓ ↓,↓,↓** (d) **↑↓ ↑↓,↑↓,↑**

(a) According Hund’s rule of maximum multiplicity, electrons will be arranged in degenerate orbitals in such a way that there multiplicity is maximum.

Multiplicity = 2|S_{T}| + 1

S_{T} → Total spin

According to Pauli’s Exclusion Principle, an orbital can have a maximum of two electrons having opposite spins.

(a) S_{T} = (1/2)+(1/2)+(1/2) = 3/2

Multiplicity = 2|3/2|+1 = 4

This case is certainly violating Pauli’s Exclusion Principle. However; it seems like that it obeys Hund’s Rule but it does not as if I rearrange the electrons according to Pauli’s Exclusion Principle, the multiplicity will drop to 2 instead of 4. Hence, this case does not obey both Pauli’s Exclusion principle as well as Hund’s rule.

In (b), only Hund’s rule is violated and (c) and (d) are correct as per both the rules.

**9) Which element has a hydrogen like spectrum whose lines have wavelength one fourth of atomic hydrogen? **

(a) **He ^{+}**

**(b)**

**Li**(c)

^{2+}**Be**(d)

^{3+}**B**

^{4+}(a) 1/λ_{H} = R×1^{2} (1/n_{1}^{2} – 1/n_{2}^{2})….(i)

1/λ_{x} = R×Z^{2} (1/n_{1}^{2} – 1/n_{2}^{2}) …..(ii)

Dividing equation (i) by equation (ii), we get

λ_{x }/λ_{H} = 1/Z^{2}

1/4 = 1/Z^{2}

Z = 2 (He^{+})

**10) The value of (n _{2} + n_{1}) and (n_{2}^{2} – n_{1}^{2}) for He^{+} ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted photon when electron jump from n_{2} to n_{1} is **

(a) **(32/9) R _{H}**

**(b)**

**(9/32) R**(c)

_{H}**9/32R**(d)

_{H}**32/9R**

_{H}(c) n_{2}+n_{1} = 4

(n_{2})^{2} – (n_{1})^{2} = 8

(n_{2} – n_{1})(n_{2}+n_{1}) = 8

(n_{2} – n_{1})(4) = 8

(n_{2} – n_{1}) = 2

n_{2} = 3 , n_{1} = 1

1/λ = R_{H}Z^{2} (1/n_{1}^{2} – 1/n_{2}^{2})

1/λ = R_{H}(2)^{2} (1/n_{1}^{2} – 1/n_{2}^{2})

1/λ = 32R_{H}/9

λ = 9/32R_{H}

**11) Which of the following statements are correct? **

(a) **The total spectral lines obtained from a single line during Zeeman effect is (2l + 1). **

(b) **In the Lyman series as the energy liberated during transition increases then the distance between the spectral lines goes on decreasing.**

(c) **The highest probability of finding an electron in 1s orbital is in the vicinity of the circumference.** ** **

(d) **The highest probability of finding an electron in 1s orbital is exactly at the middle between nucleus and circumference.**

(a, b, c)

(a) The total lines obtained due to the splitting of a spectral line in the presence of magnetic effect is (2l + 1) as the presence of orbitals which have specific orientation in the presence of external field take up certain new orientation. The number of orbitals are equal to (2l + 1) and for each orbital one splitting takes place.

(b) As we move away from the nucleus the difference in the energy levels become lesser and lesser hence.

For 1st line of Lyman ΔE = E1 (1/n_{1}^{2} – 1/n_{n}^{2})

ΔE_{1} = E_{1} (3/4) – 1^{st} line; ΔE_{2} = E_{1} (8/9) –2^{nd} line

ΔE_{3} = E_{1} (15/16) –34^{rd} line ; ΔE_{4} = E_{1} (24/25) – 4^{th} line;

ΔE_{5} = E_{1} (35/36) – 5^{th} line; ΔE_{2} – ΔE_{1} = E_{1} (8/9 – 3/4) = E_{1} (5/36)

ΔE_{3} – ΔE_{2} = E_{1} (15/16 – 8/9) = E_{1} (135-128)/144 = E_{1} (7/144)

ΔE_{4} – ΔE_{3} = E_{1} (24/25 – 15/16) = E_{1} (16×24 – 15×25)/(25×16)

= E_{1} (9/400)

We find that (ΔE_{2} – ΔE_{1}) > (ΔE_{3} – ΔE_{2}) > (ΔE_{4} – ΔE_{3})

Hence the distance also become lesser and lesser.

(c) The distance from the nucleus for maximum probability of finding electrons is 0.53 Å. This is not on the circumference of the orbital but is in the vicinity of the circumference.

**12) Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV. Which of the following results are correct about experiment?**

(a) **The velocity of photoelectrons increases with increase in intensity of radiation **

(b) **Photo-emission of electrons takes place**

(c) **the threshold frequency of the metal is 1.21 × 10 ^{15} sec^{–1} **

(d) **the velocity of the photo-electrons is 4.48 × 10 ^{6} m/sec. **

(b, c, d) According to emission photoelectric effect

hν= hν_{0} + (1/2) mV^{2}

E of incident radiation = hc/λ = (6.626×10^{-34}×3×10^{8})/200×10^{-10} = 9.94 × 10^{–18} J

ν_{0} = (5×1.6×10^{-19})/(6.623×10^{-34}) = 1.2 × 10^{15} sec^{–1}

(1/2) mv^{2} = 9.94 × 10^{–18} – 8 × 10^{–19} = 9.14 × 10^{–18}

v = √(2×9.144×10^{-18})/√(9.1×10^{-31})

= 4.47 × 10^{6} m/sec

**13) Energy equivalent of 10.00 cm ^{–1} is – **

(a) **2 × 10 ^{–22} J per photon **(b)

**2.9 × 10**

^{–2}kcal mol^{–1}photon(c) **1.2 × 10 ^{–1} kJ mol^{–1} photon **(d)

**2 × 10**

^{–15}ergs per photon(a, b, c, d) 10 cm^{–1} = 1000 m^{–1}

∴ E = hν = hc/λ = hcν^{–}

= 6.66 × 10^{–34} (Js) × 3 × 10^{8} (ms^{–1}) × 1000 (m^{–1})

= 2 × 10^{–22} J per photon = 2 × 10^{–15} ergs per photon

= 2 × 10^{–22} × 6.02 × 10^{23} J mol^{–1}

= 1.2 × 10^{–1} kJ mol^{–1}

= 2.9 × 10^{–2} kcal mol^{–1 }

**14) Identify the correct statement (s): **

(a) **Energy of the electron in 4 ^{th} orbit of H^{–}atom is more than that of the electron present in 4^{th} orbit of He^{+} ion **

(b) **Energy of a quantum is inversely proportional to frequency (v)**

(c) **Energy of a quantum is inversely proportional to wavelength (λ) **

(d) **Radius of I ^{st} orbit of He^{+} ion is 0.529 Å **

(a, c) (a) E_{H(n = 4)} = – 13.6 × (1/16) eV = –0.85 eV

E_{He+(n=4)}= –13.6 × (4/16) eV = –3.4 eV

So – 0.85 > – 3.4

⇒ E_{H(n = 4)} > E_{He+(n=4)}

(c) E = hc/λ ⇒ E ∝ 1/λ

**15) Which of the following statement (s) is/are correct? **

(a) **Fe ^{3+} and Mn^{2+} have equal paramagnetic character**

(b) **Cu _{2}Cl_{2} and CuCl_{2} are coloured.**

(c) **MnO _{4}^{–} is purple in colour because of unpair d electrons**

(d) **The magnetic moment of Fe ^{2+} and Co^{3+} both are equal to 2√6 B.M**

(a, d) Both Fe^{3+} & Mn^{2+} have 5 unpair electrons

Cu_{2}Cl_{2} is colourless because Cu^{+} has no unpair electrons MnO_{4}^{–} has no unpair electrons in Mn^{+7}, due to charge transfer spectra it is coloured.

**16) In which of the following elements Aufbau principle is/are not followed? **

(a) **Tc (43)** ** ** (b) **Ru (44)** (c) **Rh (45) ** (d) **Pd (46)**

(b, c, d) In Tc : 4 d^{5} 5s^{2} Aufbau principle is followed

In Ru : 4 d^{7} 5s^{1} Aufbau principle is not followed

In Rh : 4 d^{8} 5s^{1} Aufbau principle is not followed

In Pd : 4 d^{10} Aufbau principle is not followed

**17) Choose the correct relations on the basis of Bohr’s theory **

(a) **Velocity of an electron ∝ 1/n **(b) **Frequency of resolutions ∝ 1/n ^{3}**

(c) **Radius of orbit ∝ n ^{2}Z **(d)

**Force on an electron ∝ 1/n**

^{4}(a,b,d) Force = (kq_{1}q_{2})/r_{2}

⇒ F ∝ 1/r^{2} and Θ r ∝ n^{2}

∴ F ∝ 1/n^{4}

**18) Which of the following species have magnetic moment 4.89 B.M. **

(a) **Fe ^{+2}**

**(b)**

**Co**(c)

^{+3}**Ni**(d)

^{+4}**Mn**

^{+2 }(a, b, c) Magnetic moment = 4.89 BM, so the species must have 4 unpaired e^{–}s. In Fe^{+2}, Co^{3+}, Ni^{+4} number of unpaired e^{–}s = 4 in their 3d – subshell. Where as in Mn^{+2} there are 5 unpaired e^{–}s in 3d subshell.

**19) An electron is excited from the ground state to the n = 3 state in hydrogen atom. Which of the following statements are true?**

(a) **It takes more energy to ionize (remove) the electron from n = 3 then from the ground state **

(b) **The electron is farther from the nucleus on average in the n = 3 state then in the ground state.**

(c) **The wavelength of light emitted if the electron drops from n = 3 to n = 2 is longer than the wavelength of the light emitted it the electron falls from n = 3 to n = 1.** ** **

(d) **The first excited state corresponds to n = 3.**

(b, c)

(a) It is easier to ionize electron from 3rd orbit than ground state

(b) The radius of the third orbit is more than first orbit.

(c) The wave length of Balmer series (n_{2}→2) is more than lyman series (n_{2}→1)

(d) The first excited state compounds to n = 2.

**20) 0.53 Å is Bohr’s radius of the first orbit. In the light of the wave mechanical model, it suggests that **

(a) **the product of ψ ^{2 }and 4πr^{2} dr increases till it reaches at the distance of 0.53Aº **

(b) **Only ψ ^{2 }goes on increasing, 4πr^{2 }dr remains constant till it reaches at the distance of 0.53Aº .**

(c) **ψ ^{2 }goes on increasing, 4πr^{2 }dr goes on decreasing till it reaches at the distance of 0.53Aº.**

(d) **only 4πr ^{2 }dr goes on increasing, ψ^{2 }remains constant till it reaches at the distance of 0.53Aº . **

(a) The probability of finding 1s electron, is maximum near nucleus. On moving farther to nucleus close to 0 0.53Aº ( Bohr’s radius) ψ^{2} decreases. 4π^{2} dr , the volume element, on moving farter to nucleus increases till it reaches at Bohr’s radius and then starts to decrease.

**21) The second line of Lyman series of H coincides with 6 ^{th} line of Paschen series of an ionic species. The atomic number of ionic species is (Assuming the value of R is same for both) **

For hydrogen 1/λ = RZ^{2} (1/1 – 1/3^{2})

1/λ = 8R/9

for ionic species

1/λ = RZ^{2} (1/3^{2} – 1/9^{2})

1/λ = RZ^{2} (8/81)

8R/9 = RZ^{2} (8/81)

Z^{2} = 81/9 = 9

Z = 3

**22) The ratio of wavelength for 2 ^{nd} line of Balmer series and 1^{st} line of Lyman series, is**

(1/λ_{1}) R (1/1 – 1/2^{2})

(1/λ_{2}) = R (1/2^{2} – 1/4^{2})

On solving

λ_{1 }= 4/3 , λ_{2 }= 16/3

λ_{2}/λ_{1 }= (16/3) × (3/4)

⇒ λ_{2 }= 4λ_{1}

**23) The uncertainty in the location of circulating electron is equal to its de Broglie wavelength. The minimum percent error in its measurement of velocity under this circumstance will be approximately. **

Δx m Δv = h/4π

Δx = λ ⇒ ΔV = h/4πmλ

ΔV/V = 1/4π

% (ΔV/V) = 100×(1/4π)

= 7.96 ≅ 8

**24) A compound of vanadium ion has a magnetic moment of 1.73 B.M. Find out number of electron pair in Venadium ion. **

μ = 1.73 B.M means number of unpaired electron is one

V_{23} 1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6}, 4s^{2} 3d^{3}

V^{+4} 1s^{2}, 2s^{2} 2p^{6}, 2s^{2} 3p^{6}, 4s^{0}3d^{1}

μ = √[n(n+2)] BM

μ = √[1(1+2)] = √3 = 1.73 BM

Number of electron pair = 9

**25) An excited hydrogen atom emits light with a wavelength of 397.2 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin. (R _{H} = 1.097×10^{7}m^{-1})**

1/λ = R_{H} (1/n_{1}^{2} – 1/n_{2}^{2})

1/(397.2×10^{-19} m) = 1.097×10^{7} m^{-1} (1/2^{2} – 1/n_{2}^{2})

1/(397.2×10^{-19}×1.097×10^{7}) = 1/4 – 1/n_{2}^{2}

n_{2}^{2 }= (3.972×1.097×4)/(3.972×1.097-4) = 49

n = 7.

**26) From nth shell electron comes back to lymen series of hydrogen atom. If wavelength of hydrogen spectrum is same as wavelength for n = 4 to n = 2 transition of He ^{+} spectrum. What is the value of n for hydrogen atom. **

For H atom

1/λ_{H} = R_{H} 1^{2 }(1/(n_{low})^{2} – 1/(n_{High})^{2})

For He^{+} ion

1/λ_{He+} = R_{H} 2^{2}(1/2^{2} – 1/4^{2})

1/λ_{He+} = R_{H} 4(16-4)/64

1/λ_{He+} = R_{H} 1^{2 }(1/(n_{low})^{2} – 1/(n_{High})^{2})

n_{low }= 1

n_{High }= 2

**27) The ratio of de-Broglie wavelength of electron in 3 ^{rd} Bohrs orbit of hydrogen atom and 1^{st} orbit of hydrogen atom is ( Assuming the radius of 1^{st} orbit is r) **

nλ = 2πr for 1^{st} orbit n = 1

λ_{1} = 2πr

for 3^{rd} orbit r_{3} = 9r

3λ_{3} = 2π×9r = 18πr

λ_{3} = 6πr

λ_{3}/λ_{1 }= 6πr/2πr = 3

**28) What will be atomic number if a transition from n = 2 to n = 1 energy level would result in the emission of x-ray with λ = 3.0×10 ^{-8} m **

1/λ = R_{H}×z^{2} (1/1^{2} – 1/2^{2})

RH = 109677 cm^{-1}

1/(3×10^{-8}) = 109677×10^{2}×z^{2}×(3/4)

z^{2} = 4

z = 2

**29) Find the quantum number n corresponding to the excited state of He ^{+} ion if one transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm. **

Given λ_{2} = 30.4×10^{-7} cm

λ_{1} = 108.5×10^{-7} cm

let excited state of He^{+} be N_{2}. it comes from n_{2} to n_{1} and then n_{1} to 1 to emit two successive photon

1/λ_{2} = R_{H}.z^{2} (1/1^{2} – 1/n_{1}^{2})

1/(30.4×10^{-7}) = 109678×4 (1/1^{2} – 1/n_{1}^{2})

n_{1} = 2

Now 1/λ_{1} = R_{H}.z^{2} (1/2^{2} – 1/n_{2}^{2})

1/(108.5×10^{-7}) = 109678×4 (1/2^{2} – 1/n_{2}^{2})

n_{2} = 5

Thus, excited state for He is 5th orbit.

**30) Hydrogen like atom with atomic no ‘Z’ is in a higher excited state of quantum no. ‘n’. This excited state can make a transition to the first excited state by successively emitting two photons of energies 20ev & 28.3 ev respectively. Alternatively the atom from the same excited state can make transition to the 2 ^{nd} excited state by emitting photons of energies 8.0ev & 10.13ev respectively. Determine ‘Z’ . **

Energy difference between, n = 2 & n = 3

= 20 + 28.3 -8.0 -10.13

= 30. 17

30. 17 = 13. 6 (1/2^{2} – 1/3^{2})×Z^{2}

⇒ z ≈ 3.99 = 4.0

Energy difference between n_{1} = n & n_{2} = 2

= 20 + 28.3 =48.3

48.3= 13.6 (1/2^{2} – 1/n^{2})4×4

n ≈ 6.0