1. The nucleus of an atom is located at \( x=y=z=0 \). If the probability of finding an s-orbital electron in a tiny volume around \( x=a, y=z=0 \) is \( 1 \times 10^{-5} \), what is the probability of finding the electron in the same sized volume around \( x=z=0, y=a \)?

• (a) \( 1 \times 10^{-5} \)
• (b) \( 1 \times 10^{-5} \times a \)
• (c) \( 1 \times 10^{-5} \times a^2 \)
• (d) \( 1 \times 10^{-5} \times a^{-1} \)

Correct Answer: (a)

Explanation: Since the distance from the nucleus is the same (\( r=a \)), and the s-orbital is spherically symmetrical, the probability density is identical in all directions at that distance.

2. The limiting line in Balmer series will have a frequency of:

• (a) \( 6.22 \times 10^{15} s^{-1} \)
• (b) \( 7.22 \times 10^{14} s^{-1} \)
• (c) \( 8.22 \times 10^{14} s^{-1} \)
• (d) \( 9.22 \times 10^{14} s^{-1} \)

Correct Answer: (c)

Explanation: The limiting line refers to the transition from \( n_2 = \infty \) to \( n_1 = 2 \).
\[ \nu = 3.29 \times 10^{15} \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right) = 3.29 \times 10^{15} \left( \frac{1}{4} – 0 \right) = 8.22 \times 10^{14} s^{-1} \].

3. If the electron falls from \( n=5 \) to \( n=4 \) in the H-atom, then emitted energy is:

• (a) 0.306 eV
• (b) 12.09 eV
• (c) 1.89 eV
• (d) 0.65 eV

Correct Answer: (a)

Explanation: \( \Delta E = 13.6 \left( \frac{1}{4^2} – \frac{1}{5^2} \right) eV = 13.6 \left( \frac{1}{16} – \frac{1}{25} \right) = 0.306 eV \).

4. Photoelectric effect is the phenomenon in which:

• (a) photons come out of a metal when hit by electrons
• (b) photons come out of the nucleus under an electric field
• (c) electrons come out with constant velocity depending on frequency
• (d) electrons come out with different velocities not greater than a certain value depending only on frequency.

Correct Answer: (d)

Explanation: According to Einstein’s equation, \( \frac{1}{2}mv_{max}^2 = h\nu – W \). The maximum velocity depends on the frequency of incident light, not intensity.

5. Which of the following statements is not correct?

• (a) Shape depends on azimuthal quantum number.
• (b) Orientation depends on magnetic quantum number.
• (c) Energy in multi-electron atoms depends only on principal quantum number.
• (d) Number of degenerate orbitals depends on \( l \) and \( m \).

Correct Answer: (c)

Explanation: In multi-electron systems, the energy level of an orbital depends upon both \( n \) and \( l \) (\( n+l \) rule).

6. Rydberg is:

• (a) also called Rydberg constant and is a universal constant
• (b) unit of wavelength
• (c) unit of wave number
• (d) unit of energy and one Rydberg equals 13.6 eV.

Correct Answer: (a)

Explanation: Rydberg refers to the Rydberg constant, which is used to calculate the wavelengths of spectral lines.

7. The total number of orbitals in a shell with principal quantum number n is:

• (a) 2n
• (b) \( 2n^2 \)
• (c) \( n^2 \)
• (d) n+1

Correct Answer: (c)

Explanation: The number of orbitals in any principal shell is given by \( n^2 \).

8. The spectrum of white light ranging from red to violet is called a continuous spectrum because:

• (a) different colours are seen as different bands
• (b) colours continuously absorb energy
• (c) the violet colour merges into blue, blue into green, etc.
• (d) it is a continuous band of coloured and white light.

Correct Answer: (c)

Explanation: In a continuous spectrum, the colours merge into each other in a continuous pattern without any sharp boundaries.

9. Effective nuclear charge (\( Z_{eff}e \)) for a nucleus of an atom is defined as:

• (a) shielding of outermost electrons
• (b) the net positive charge experienced by electron from the nucleus
• (c) attractive force experienced by nucleus
• (d) screening of positive charge.

Correct Answer: (b)

Explanation: It is the net positive charge felt by an electron, which is less than the actual nuclear charge due to shielding by inner electrons.

10. The number of radial nodes and angular nodes for d-orbital can be represented as:

• (a) \( (n-2) \) radial nodes + 1 angular node
• (b) \( (n-1) \) radial nodes + 1 angular node
• (c) \( (n-3) \) radial nodes + 2 angular nodes \( = (n-l-1) \) total nodes
• (d) \( (n-3) \) radial nodes + 2 angular nodes \( = (n-1) \) total nodes.

Correct Answer: (d)

Explanation: For a d-orbital, \( l=2 \). Angular nodes \( = l = 2 \). Radial nodes \( = n – l – 1 = n – 3 \). Total nodes \( = n – 1 \).

11. If the radius of first Bohr orbit is x pm, then the radius of the third orbit would be:

• (a) \( 3x \) pm
• (b) \( 6x \) pm
• (c) \( 9x \) pm
• (d) \( \frac{1}{2}x \) pm

Correct Answer: (c)

Explanation: Radius \( r \propto n^2 \). For \( n=3 \), \( r = x \times 3^2 = 9x \) pm.

12. Millikan’s oil drop method is used to find:

• (a) e/m ratio of electron
• (b) mass of electron
• (c) velocity of electron
• (d) charge of electron.

Correct Answer: (d)

Explanation: This method was historically used to determine the fundamental unit of charge on an electron.

13. The correct set of quantum numbers for the outermost electron of Rubidium (37) is:

• (a) 5, 0, 0, \( +\frac{1}{2} \)
• (b) 4, 3, 2, \( -\frac{1}{2} \)
• (c) 5, 1, 0, \( -\frac{1}{2} \)
• (d) 5, 1, 1, \( +\frac{1}{2} \)

Correct Answer: (a)

Explanation: Rubidium (At. No. 37) is an alkali metal in the 5th period. Its outermost electron is in the 5s subshell. Thus, \( n=5, l=0, m=0, s=\pm \frac{1}{2} \).

14. In any subshell, the maximum number of electrons having same value of spin quantum number is:

• (a) \( \sqrt{l(l+1)} \)
• (b) \( l+2 \)
• (c) \( 2l+1 \)
• (d) \( 4l+2 \)

Correct Answer: (c)

Explanation: The maximum number of electrons with the same spin is equal to the number of orbitals in the subshell, which is \( 2l+1 \).

15. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. The ion is:

• (a) \( {}_{28}^{56}Ni^{3+} \)
• (b) \( {}_{26}^{56}Fe^{3+} \)
• (c) \( {}_{27}^{56}Co^{3+} \)
• (d) \( {}_{24}^{56}Cr^{3+} \)

Correct Answer: (b)

Explanation: Let \( e^- = x \). Neutrons \( = 1.304x \). Protons \( = x+3 \). Mass No. \( = (x+3) + 1.304x = 56 \). Solving gives \( x=23 \). Protons \( = 26 \) (Iron).

16. Energy of H-atom in the ground state is -13.6 eV, hence energy in the second excited state is:

• (a) -6.8 eV
• (b) -3.4 eV
• (c) -1.51 eV
• (d) -4.53 eV

Correct Answer: (c)

Explanation: Second excited state corresponds to \( n=3 \).
\[ E_3 = -\frac{13.6}{3^2} = -1.51 eV \].

17. The total spin resulting from a \( d^7 \) configuration is:

• (a) \( \pm 1/2 \)
• (b) \( \pm 2 \)
• (c) \( \pm 1 \)
• (d) \( \pm 3/2 \)

Correct Answer: (d)

Explanation: A \( d^7 \) subshell has 3 unpaired electrons. Total spin \( = \pm \frac{1}{2} \times 3 = \pm \frac{3}{2} \).

18. Nuclides:

• (a) have same number of protons
• (b) have specific atomic numbers
• (c) have specific atomic and mass numbers
• (d) are isotopes.

Correct Answer: (c)

Explanation: A nuclide is an atomic species characterized by the specific constitution of its nucleus, i.e., by its number of protons and neutrons.

19. \( {}^{18}O \) isotope of oxygen will have:

• (a) 18 protons
• (b) 9 protons and 9 neutrons
• (c) 8 neutrons and 10 protons
• (d) 10 neutrons and 8 protons.

Correct Answer: (d)

Explanation: Atomic number of Oxygen is 8. For mass number 18, neutrons \( = 18 – 8 = 10 \).

20. The concept that atoms combine in small whole number ratio was proposed by:

• (a) Dalton
• (b) Avogadro
• (c) Gay-Lussac
• (d) Berzelius.

Correct Answer: (a)

Explanation: This is a core postulate of Dalton’s Atomic Theory.

21. For a shell of principal quantum number \( n=4 \), there are:

• (a) 16 orbitals
• (b) 8 subshells
• (c) 18 electrons (maximum)
• (d) 4 electrons with \( l=3 \).

Correct Answer: (a)

Explanation: Maximum orbitals \( = n^2 = 4^2 = 16 \). Max electrons would be \( 2n^2 = 32 \).

22. Which among the following statements is not correct?

• (a) \( \Psi^2 \) represents the molecular orbitals.
• (b) The number of peaks in radial distribution is \( (n-l) \).
• (c) Radial probability density \( \rho_{nl}(r) = 4 \pi r^2 R^2_{nl}(r) \).
• (d) A node has zero amplitude.

Correct Answer: (a)

Explanation: \( \Psi^2 \) represents the atomic orbital (probability density at a point).

23. What is the uncertainty involved in the measurement of velocity of an electron located within a distance of 0.1 Å?

• (a) \( 5.79 \times 10^6 cm s^{-1} \)
• (b) \( 5.79 \times 10^6 m s^{-1} \)
• (c) \( 5.79 \times 10^{10} cm s^{-1} \)
• (d) \( 5.79 \times 10^{10} m s^{-1} \)

Correct Answer: (b)

Explanation: Using \( \Delta x \cdot m \Delta v = \frac{h}{4 \pi} \). With \( \Delta x = 10^{-11} m \), \( \Delta v = 5.79 \times 10^6 m/s \).

24. A golf ball has a mass of 40 g and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.

• (a) \( 1.46 \times 10^{-33} m \)
• (b) \( 1.46 \times 10^{-33} cm \)
• (c) \( 1.59 \times 10^{-33} m \)
• (d) \( 1.39 \times 10^{33} km \)

Correct Answer: (a)

Explanation: \( \Delta v = 45 \times 0.02 = 0.9 m/s \).
\[ \Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.626 \times 10^{-34}}{4 \times \pi \times 0.04 \times 0.9} = 1.46 \times 10^{-33} m \].

25. Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length \( 1 \times 10^{-10} \) metre.

• (a) \( 5.37 \times 10^{-27} kg m s^{-1} \)
• (b) \( 5.27 \times 10^{-25} g m s^{-1} \)
• (c) \( 5.37 \times 10^{-25} g m s^{-1} \)
• (d) \( 5.27 \times 10^{-25} kg m s^{-1} \)

Correct Answer: (d)

Explanation: \( \Delta p = \frac{h}{4 \pi \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times \pi \times 10^{-10}} = 5.27 \times 10^{-25} kg m s^{-1} \).

26. What is the maximum number of electrons that can be associated with the following set of quantum numbers? \( n=3, l=1 \) and \( m=-1 \)

• (a) 4
• (b) 2
• (c) 10
• (d) 6

Correct Answer: (b)

Explanation: The quantum numbers define a specific orbital (3p). Any single orbital can accommodate a maximum of 2 electrons.

27. Based on equation \( E = -2.178 \times 10^{-18} J \left( \frac{Z^2}{n^2} \right) \), which conclusion is not correct?

• (a) Equation can calculate energy change during orbit shifts.
• (b) For \( n=1 \), the electron is more loosely bound than for \( n=6 \).
• (c) The negative sign means bound energy is lower than at infinite distance.
• (d) Larger n implies larger orbit radius.

Correct Answer: (b)

Explanation: For \( n=1 \), the energy is most negative, meaning the electron is more tightly bound to the nucleus.

28. The ionization enthalpy of \( He^+ \) ion is \( 19.60 \times 10^{-18} J/atom \). The ionization enthalpy of \( Li^{2+} \) ion will be:

• (a) \( 84.2 \times 10^{-18} J/atom \)
• (b) \( 44.10 \times 10^{-18} J/atom \)
• (c) \( 63.20 \times 10^{-18} J/atom \)
• (d) \( 21.20 \times 10^{-18} J/atom \)

Correct Answer: (b)

Explanation: \( I.E. \propto Z^2 \).
\[ I.E.(Li^{2+}) = I.E.(He^+) \times \frac{3^2}{2^2} = 19.6 \times \frac{9}{4} = 44.1 \times 10^{-18} J/atom \].

29. Which transition in the hydrogen atomic spectrum will have the same wavelength as the transition \( n=4 \) to \( n=2 \) of \( He^+ \) spectrum?

• (a) \( n=4 \) to \( n=3 \)
• (b) \( n=3 \) to \( n=2 \)
• (c) \( n=4 \) to \( n=2 \)
• (d) \( n=2 \) to \( n=1 \)

Correct Answer: (d)

Explanation: For \( He^+ (Z=2) \), \( \frac{1}{\lambda} = 4R \left( \frac{1}{4} – \frac{1}{16} \right) = \frac{3R}{4} \). For H, \( \frac{1}{\lambda} = R \left( \frac{1}{1} – \frac{1}{4} \right) = \frac{3R}{4} \).

30. The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The possible energy value of electron in the excited state of \( Li^{2+} \) is:

• (a) -122.4 eV
• (b) 30.6 eV
• (c) -30.6 eV
• (d) 13.6 eV

Correct Answer: (c)

Explanation: For \( Li^{2+} (Z=3) \), \( E_n = -13.6 \frac{Z^2}{n^2} \). For \( n=2 \), \( E = -13.6 \times \frac{9}{4} = -30.6 eV \).

31. Which one of the following has the lowest ionisation energy?

• (a) \( 1s^2 2s^2 2p^6 \)
• (b) \( 1s^2 2s^2 2p^6 3s^1 \)
• (c) \( 1s^2 2s^2 2p^5 \)
• (d) \( 1s^2 2s^2 2p^3 \)

Correct Answer: (b)

Explanation: The structure \( 3s^1 \) corresponds to Sodium, an alkali metal. Alkali metals have the lowest ionization energies in their periods.

32. The representation of the ground state electronic configuration of He as ↑↑ is wrong because it violates:

• (a) Heisenberg’s uncertainty principle
• (b) Bohr’s quantization theory
• (c) Pauli’s exclusion principle
• (d) Hund’s rule.

Correct Answer: (c)

Explanation: Pauli’s principle states an orbital can have maximum two electrons with opposite spins.

33. The electronic transitions from \( n=2 \) to \( n=1 \) will produce shortest wavelength in:

• (a) \( Li^{2+} \)
• (b) \( He^+ \)
• (c) H
• (d) \( H^+ \)

Correct Answer: (a)

Explanation: \( \lambda \propto \frac{1}{Z^2} \). Since Lithium has the highest atomic number (Z=3) among the choices, it produces the shortest wavelength.

34. 13.6 eV is needed for ionization of a hydrogen atom. An electron absorbs 1.50 times the escaping energy. What is its wavelength?

• (a) \( 1.5 \times 10^{-6} m \)
• (b) \( 2.3 \times 10^{-11} m \)
• (c) \( 4.7 \times 10^{-10} m \)
• (d) \( 1.4 \times 10^{-9} m \)

Correct Answer: (c)

Explanation: Energy absorbed \( = 20.4 eV \). KE \( = 20.4 – 13.6 = 6.8 eV \).
\[ v = \sqrt{\frac{2KE}{m}}; \lambda = \frac{h}{mv} = 4.7 \times 10^{-10} m \].

35. Number of waves made by an electron in one complete revolution in 3rd Bohr orbit is:

• (a) 2
• (b) 3
• (c) 4
• (d) 1

Correct Answer: (b)

Explanation: The number of waves in an orbit is equal to the principal quantum number \( n \). For 3rd orbit, \( n=3 \).

36. The degeneracy of the level of hydrogen atom that has energy \( -\frac{R_H}{16} \) is:

• (a) 16
• (b) 4
• (c) 2
• (d) 1

Correct Answer: (a)

Explanation: \( E = -\frac{R_H}{n^2} \). Here \( n^2 = 16 \), so \( n=4 \). Degeneracy \( = n^2 = 16 \).

37. What is the minimum error in position of an electron moving with speed 500 m/s measured to accuracy of 0.006%?

• (a) \( 2.73 \times 10^{-3} m \)
• (b) \( 1.932 \times 10^{-3} m \)
• (c) \( 3.112 \times 10^{-5} m \)
• (d) \( 0.119 \times 10^{-5} m \)

Correct Answer: (b)

Explanation: \( \Delta v = 500 \times \frac{0.006}{100} = 0.03 m/s \).
\[ \Delta x = \frac{h}{4 \pi m \Delta v} = 1.932 \times 10^{-3} m \].

38. If the photon detector receives a total of \( 3.15 \times 10^{-18} J \) from radiation of 600 nm, calculate the number of photons.

• (a) 4
• (b) 9
• (c) 6
• (d) 7

Correct Answer: (b)

Explanation: \( E_{total} = n \left( \frac{hc}{\lambda} \right) \).
\[ n = \frac{3.15 \times 10^{-18} \times 600 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} \approx 9.5 \approx 9 \].

39. Time taken for an electron to complete one revolution in the Bohr orbit is:

• (a) \( \frac{4 \pi^2 m r^2}{nh} \)
• (b) \( \frac{nh}{4 \pi^2 m r} \)
• (c) \( \frac{2 \pi m r}{n^2 h^2} \)
• (d) \( \frac{nh}{4 \pi^2 m r^2} \)

Correct Answer: (a)

Explanation: \( t = \frac{2 \pi r}{v} \). Since \( v = \frac{nh}{2 \pi m r} \), substituting gives \( t = \frac{4 \pi^2 m r^2}{nh} \).

40. What is the wavelength of light emitted when electron undergoes transition from \( n=4 \) to \( n=2 \) in H-atom?

• (a) 205 nm
• (b) 123 nm
• (c) 486 nm
• (d) 329 nm

Correct Answer: (c)

Explanation: Using Rydberg formula:
\[ \bar{\nu} = 109677 \left( \frac{1}{4} – \frac{1}{16} \right) = 20564 cm^{-1}; \lambda = 486 nm \].

41. The energy of the first electron in helium will be:

• (a) -13.6 eV
• (b) -54.4 eV
• (c) -5.44 eV
• (d) zero

Correct Answer: (b)

Explanation: For \( He^+ \), \( E = -13.6 \times \frac{Z^2}{n^2} = -13.6 \times 4 = -54.4 eV \).

42. Calculate value of n for Paschen transition (\( n_1=3 \)) observed at 1285 nm.

• (a) 3
• (b) 4
• (c) 5
• (d) 6

Correct Answer: (c)

Explanation: Using the given formula \( \nu = 3.29 \times 10^{15} \left( \frac{1}{9} – \frac{1}{n^2} \right) \), solving for \( \lambda = 1285 nm \) gives \( n=5 \).

43. If the uncertainty in the position of an electron is zero, the uncertainty in its momentum would be:

• (a) zero
• (b) greater than \( h/4 \pi \)
• (c) less than \( h/4 \pi \)
• (d) infinite.

Correct Answer: (d)

Explanation: \( \Delta p = \frac{h}{4 \pi \Delta x} = \frac{h}{0} = \infty \).

44. If position is known within \( 10^{-12} m \), the uncertainty in momentum is:

• (a) \( 5.27 \times 10^{-23} kg m s^{-1} \)
• (b) \( 2.130 \times 10^{-23} kg m s^{-1} \)
• (c) \( 1.69 \times 10^{-24} kg m s^{-1} \)
• (d) \( 1.69 \times 10^{-23} kg m s^{-1} \).

Correct Answer: (a)

Explanation: \( \Delta p = \frac{6.626 \times 10^{-34}}{4 \pi \times 10^{-12}} = 5.27 \times 10^{-23} kg m s^{-1} \).

45. Order of increasing energy from lowest to highest: (i) \( n=4, l=1 \); (ii) \( n=4, l=0 \); (iii) \( n=3, l=2 \); (iv) \( n=3, l=1 \).

• (a) \( (iv) < (ii) < (iii) < (i) \)
• (b) \( (ii) < (iv) < (i) < (iii) \)
• (c) \( (i) < (iii) < (ii) < (iv) \)
• (d) \( (iii) < (i) < (iv) < (ii) \)

Correct Answer: (a)

Explanation: Using the \( (n+l) \) rule: (iv) \( 3p \implies 4 \); (ii) \( 4s \implies 4 \); (iii) \( 3d \implies 5 \); (i) \( 4p \implies 5 \). Between same \( n+l \), lower \( n \) has lower energy. Thus \( 3p < 4s < 3d < 4p \).