Question 1

Consider two radiations of wavelengths 1. _1 = 2000 Å 2. _2 = 6000 Å The ratio of the energies of these two radiations ( E_1 E_2 ) is ________ (Nearest integer).

Accepted Answer

The energy of a photon is given by the relation: E = hc For the two radiations, the energies are E_1 = hc _1 and E_2 = hc _2 . The ratio of their energies is: E_1 E_2 = _2 _1 Substituting the given values _1 = 2000 and _2 = 6000: E_1 E_2 = 6000 2000 = 3 Answer: 3

Question 2

The Bohr radius of a hydrogen like species is 70.53 pm. The species and the stationary state (n) are respectively (Given : Hydrogen atom Bohr radius is 52.9 pm)

Accepted Answer

The radius of the n-th Bohr orbit for a hydrogen-like species is given by the formula: r_n = a_0 n^2 Z where a_0 is the Bohr radius of the hydrogen atom (52.9 pm), n is the principal quantum number, and Z is the atomic number. Given r_n = 70.53 pm, we can write: 70.53 = 52.9 n^2 Z n^2 Z = 70.53 52.9 = 1.333 = 4 3 Checking the given options: For Li ^ 2+ and n = 3, Z = 3 n^2 Z = 9 3 = 3 For He ^+ and n = 3, Z = 2 n^2 Z = 9 2 = 4.5 For He ^+ and n = 2, Z = 2 n^2 Z = 4 2 = 2 For Li ^ 2+ and n = 2, Z

Question 3

If shortest wavelength of hydrogen atom in Lyman series is x, then longest wavelength in Balmer series of He ^+ is:

Accepted Answer

For the shortest wavelength in the Lyman series of the hydrogen atom (Z=1), the transition is from n_2 = to n_1 = 1. 1 x = R(1)^2 ( 1 1^2 - 1 ^2 ) = R This gives R = 1 x . For the longest wavelength in the Balmer series of He ^+ (Z=2), the transition is from n_2 = 3 to n_1 = 2. 1 = R(2)^2 ( 1 2^2 - 1 3^2 ) 1 = 4R ( 1 4 - 1 9 ) 1 = 4R ( 5 36 ) = 5R 9 Substituting R = 1 x : 1 = 5 9x = 9x 5 Answer: 9x 5

Question 4

Match the LIST-I with LIST-II List-I Orbital List-II Radial nodes and nodal plane A. 2s I. 1 Radial node + two nodal planes B. 3s II. 1 Radial node + one nodal plane C. 3p III. 2 Radial nodes + No nod

Accepted Answer

The number of radial nodes is given by the formula n - l - 1. The number of nodal planes (angular nodes) is given by l. For 2s orbital: n = 2, l = 0. Radial nodes = 2 - 0 - 1 = 1. Nodal planes = 0. This matches A with IV. For 3s orbital: n = 3, l = 0. Radial nodes = 3 - 0 - 1 = 2. Nodal planes = 0. This matches B with III. For 3p orbital: n = 3, l = 1. Radial nodes = 3 - 1 - 1 = 1. Nodal planes = 1. This matches C with II. For 4d orbital: n = 4, l = 2. Radial nodes = 4 - 2 - 1 = 1. Nodal planes

Question 5

Which of the following statement(s) is/are true ? A. If two orbitals have the same value of (n + l), the orbital with lower value of n will have lower energy. B. Energies of the orbitals in the same s

Accepted Answer

Statement A: According to the (n+l) rule, if two orbitals have the same (n+l) value, the orbital with the lower value of n has lower energy. This statement is true. Statement B: As the atomic number increases, the effective nuclear charge (Z_ eff ) increases. The energy of an orbital is proportional to - Z_ eff ^2 n^2 . Thus, the energy becomes more negative, meaning the energy of the orbitals in the same subshell decreases with an increase in atomic number. This statement is false. Statement C:

Question 6

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. A. n = 3, l = 2, m = +1 B. n = 4, l = 0, m = 0 C. n = 6, l = 1, m = 0 D. n = 5, l = 1, m = +1 E. n = 2, l =

Accepted Answer

The energy of an atomic orbital in a multi-electron atom is determined by the (n + l) rule. Orbitals with a lower value of (n + l) have lower energy. If two orbitals have the same (n + l) value, the one with the lower n value has lower energy. Calculating the (n + l) value for each given orbital: A. n = 3, l = 2 n + l = 5 B. n = 4, l = 0 n + l = 4 C. n = 6, l = 1 n + l = 7 D. n = 5, l = 1 n + l = 6 E. n = 2, l = 1 n + l = 3 Arranging the orbitals in increasing order of their (n + l) values, we g

Question 7

Identify the correct statements from the following: A. Heisenberg uncertainty principle is applicable to electrons. B. The size of 2p_x orbital is less than the size of 3p_x orbital. C. The energy of

Accepted Answer

Statement A is correct. Heisenberg's uncertainty principle is applicable to microscopic particles such as electrons. Statement B is correct. The size of an orbital increases with an increase in the principal quantum number n. Thus, the size of the 3p_x orbital is greater than that of the 2p_x orbital. Statement C is incorrect. The energy of an orbital depends on the effective nuclear charge. Since the nuclear charge of Li (Z=3) is greater than that of H (Z=1), the energy of the 2s orbital of Li

Question 8

The species having identical radii according to the Bohr's theory are: A. H (first orbit) B. He^+ (first orbit) C. He^+ (Second orbit) D. Li^ 2+ (first orbit) E. Be^ 3+ (Second orbit) Choose the corre

Accepted Answer

According to Bohr's theory, the radius of the n-th orbit of a hydrogen-like species is given by r_n = 0.529 n^2 Z . For H (first orbit), n = 1, Z = 1 r = 0.529 1^2 1 = 0.529 . For He^+ (first orbit), n = 1, Z = 2 r = 0.529 1^2 2 = 0.2645 . For He^+ (second orbit), n = 2, Z = 2 r = 0.529 2^2 2 = 1.058 . For Li^ 2+ (first orbit), n = 1, Z = 3 r = 0.529 1^2 3 = 0.176 . For Be^ 3+ (second orbit), n = 2, Z = 4 r = 0.529 2^2 4 = 0.529 . The radii of H (first orbit) and Be^ 3+ (second orbit) are identi

Question 9

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

Accepted Answer

The wave number of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: = R_H ( 1 n_1^2 - 1 n_2^2 ) For the first line (lowest energy) of the Balmer series, the transition is from n_2 = 3 to n_1 = 2: _ Balmer = R_H ( 1 2^2 - 1 3^2 ) = R_H ( 1 4 - 1 9 ) = 5R_H 36 For the first line of the Brackett series, the transition is from n_2 = 5 to n_1 = 4: _ Brackett = R_H ( 1 4^2 - 1 5^2 ) = R_H ( 1 16 - 1 25 ) = 9R_H 400 Taking the ratio of the two wave numbers: _ Balmer _

Question 10

Which of the following is correct set of 4 quantum numbers of 19^ th electron in Chromium (Atomic number = 24) in accordance with Aufbau principle?

Accepted Answer

The atomic number of Chromium is 24. According to the Aufbau principle, the order of filling of electrons is 1s, 2s, 2p, 3s, 3p, 4s, 3d. The electronic configuration of the first 18 electrons is 1s^2 2s^2 2p^6 3s^2 3p^6. The 19^ th electron enters the 4s orbital. For the 4s orbital, the quantum numbers are: Principal quantum number, n = 4 Azimuthal quantum number, l = 0 Magnetic quantum number, m = 0 Spin quantum number, s = + 1 2 or - 1 2 This matches the set given in the fourth option. Answer:

Question 11

A monoatomic anion (A^-) has 45 neutrons and 36 electrons. Atomic mass, group in the periodic table and physical state at room temperature of the element (A) respectively are

Accepted Answer

Number of electrons in A^ - is 36. Number of electrons in neutral atom A = 36 - 1 = 35. Atomic number of A = 35. Number of protons in A = 35. Atomic mass of A = Number of protons + Number of neutrons = 35 + 45 = 80. The element with atomic number 35 is Bromine ( Br ). Bromine belongs to Group 17 of the periodic table. The physical state of Bromine at room temperature is liquid. Answer: 80, 17, liquid

Question 12

The surface of sodium metal is irradiated with radiation of wavelength x nm. The kinetic energy of ejected electrons is 2.8 10^ -20 J. The work function of sodium is 2.3 eV. The value of x is _____ 10

Accepted Answer

The work function of sodium in Joules is: = 2.3 1.6 10^ -19 = 3.68 10^ -19 J The kinetic energy of the ejected electrons is: K = 2.8 10^ -20 = 0.28 10^ -19 J Using the photoelectric equation, the energy of the incident photon is: E = + K E = 3.68 10^ -19 + 0.28 10^ -19 = 3.96 10^ -19 J The wavelength of the incident radiation is calculated using E = hc : = hc E = 6.6 10^ -34 3.0 10^8 3.96 10^ -19 = 19.8 10^ -26 3.96 10^ -19 = 5 10^ -7 m Converting the wavelength to nanometers: = 500 10^ -9 m = 5

Question 13

What is the energy (in J atom^ -1 ) required for the following process? Li^ 2+ (g) Li^ 3+ (g) + e^- (Take the ionization energy for the H atom in the ground state as 2.18 10^ -18 J atom^ -1 )

Accepted Answer

The given process Li^ 2+ (g) Li^ 3+ (g) + e^- represents the ionization of the hydrogen-like ion Li^ 2+ from its ground state. The ionization energy for a hydrogen-like species is given by the formula: IE = IE_ H Z^2 For Li^ 2+ , the atomic number Z = 3. Given the ionization energy of the hydrogen atom IE_ H = 2.18 10^ -18 J atom^ -1 . Substituting the values: IE = 2.18 10^ -18 (3)^2 IE = 2.18 10^ -18 9 IE = 19.62 10^ -18 J atom^ -1 IE = 1.962 10^ -17 J atom^ -1 Answer: 1.962 10^ -17

Question 14

The wavelength of photon ' A ' is 400 nm. The frequency of photon ' B ' is 10^ 16 ~s ^ -1 . The wave number of photon ' C^ is 10^ 4 ~cm ^ -1 . The correct order of energy of these photons is :

Accepted Answer

Photon A: _A = 400 nm E_A = hc _A = 6.626 10^ -34 3 10^8 400 10^ -9 = 4.97 10^ -19 J Photon B: _B = 10^ 16 s⁻¹ E_B = h _B = 6.626 10^ -34 10^ 16 = 6.626 10^ -18 J Photon C: _C = 10^4 cm⁻¹ = 10^6 m⁻¹ _C = 10^ -6 m E_C = hc _C = 6.626 10^ -34 3 10^8 10^ -6 = 1.988 10^ -19 J Order: E_B > E_A > E_C

Question 15

Two positively charged particles m _ 1 and m _ 2 have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m _ 1 =1 amu and m _ 2 =4 amu] The deBroglie wavel

Accepted Answer

For a charged particle accelerated through potential V, the de Broglie wavelength is = h 2mVe . The wavelength is inversely proportional to m . Therefore: _1 _2 = m_2 m_1 = 4 1 = 2. This means _1 = 2 _2, so x = 2 (nearest integer).

Question 16

The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral lines belonging to Balmer series. ( R = Rydberg constant)

Accepted Answer

For the Balmer series, transitions end at n=2. Using the Rydberg formula = R ( 1 2^2 - 1 n^2 ) = R ( 1 4 - 1 n^2 ), we calculate wavenumbers for transitions from different upper levels: From n=3: = R ( 1 4 - 1 9 ) = 5R 36 From n=4: = R ( 1 4 - 1 16 ) = 3R 16 From n=5: = R ( 1 4 - 1 25 ) = 21R 100 Option (4) contains exactly these three wavenumbers corresponding to the Balmer series transitions.

Question 17

Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?

Accepted Answer

The 2s orbital has one radial nodal surface, which is a spherical surface shown in Figure 1. A nodal surface is where the electron probability density (and wave function) equals zero, meaning _ 2s (x) = 0. From the wave function graph in Figure 2, point B represents where the wave function crosses zero on the x-axis, indicating the location of the radial node. Points A and C show maximum and minimum values respectively, while point D shows a non-zero region beyond the node.

Question 18

The wavelength of spectral line obtained in the spectrum of Li ^ 2+ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

Accepted Answer

For Li²⁺ (Z=3): From n_2 + n_1 = 4 and n_2 - n_1 = 2, we get n_2 = 3, n_1 = 1 Using Bohr wavelength formula: 1 = RZ^2 ( 1 n_1^2 - 1 n_2^2 ) 1 = 1.097 10^7 9 (1 - 1 9 ) = 9.873 10^7 8 9 1 = 8.776 10^7 m⁻¹ = 1.14 10^ -6 cm

Question 19

The hydrogen spectrum consists of several spectral lines in Lyman series ( L _ 1 , ~L _ 2 , L _ 3 ; L _ 1 has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balm

Accepted Answer

The energy of a spectral line in the hydrogen spectrum is given by E = 13.6 ( 1 n_1^2 - 1 n_2^2 ) eV. For the Lyman series, the lowest energy line L_1 corresponds to the transition from n=2 to n=1. E(L_1) = 13.6 ( 1 1^2 - 1 2^2 ) = 13.6 ( 1 - 1 4 ) = 13.6 3 4 eV. For the Balmer series, the lowest energy line B_1 corresponds to the transition from n=3 to n=2. E(B_1) = 13.6 ( 1 2^2 - 1 3^2 ) = 13.6 ( 1 4 - 1 9 ) = 13.6 5 36 eV. Given E(L_1) = x E(B_1), we have: 13.6 3 4 = x 13.6 5 36 3 4 = x 5 36

Question 20

The work functions of two metals (M_ A . and .M_ B ) are in the 1: 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of M_ A : M_ B is in the

Accepted Answer

The work functions satisfy W_A : W_B = 1:2. Let W_A = W and W_B = 2W. Using the photoelectric equation KE = h - W, we have KE_A = 6 - W and KE_B = 6 - 2W. Given the kinetic energy ratio KE_A KE_B = 2.642, we get 6-W 6-2W = 2.642. Solving: 6 - W = 2.642(6 - 2W) gives 6 - W = 15.852 - 5.284W So 4.284W = 9.852 and W = 2.3 eV. Thus W_A = 2.3 eV and W_B = 4.6 eV.

Structure of Atom — Previous Year Questions & Solutions