JEE Chemistry Structure Of Atom MCQs Set 06

Chemistry Objective Questions and Answers: Structure Of Atom

Access targeted multiple-choice questions for Structure Of Atom designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: The charge to mass ratio of the particles forming cathode rays, for different gases in the discharge tube,
(a) is largest for hydrogen
(b) is smallest for hydrogen
(c) depends on the metal of the cathode only
(d) is the same for all gases
Show Answer & Explanation

Answer: (d) is the same for all gases

Explanation:
1. Cathode rays are electrons
2. Every atom contains the same electron, so \(e/m = 1.76\times10^{11}\) C kg\(^{-1}\) whatever the gas
3. Only anode rays show a gas-dependent e/m.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 2: The fraction of the volume of an atom that is occupied by its nucleus is of the order of
(a) \(10^{-5}\)
(b) \(10^{-30}\)
(c) \(10^{-15}\)
(d) \(10^{-10}\)
Show Answer & Explanation

Answer: (c) \(10^{-15}\)

Explanation:
1. Nuclear radius is about \(10^{-13}\) cm and atomic radius about \(10^{-8}\) cm, so the volume ratio is \((10^{-5})^3 = 10^{-15}\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 3: The radius of a hydrogen atom is 0.05 nm and the radius of a proton is \(1.5\times10^{-15}\) m. The fraction of the atomic volume occupied by the nucleus is nearest to
(a) \(3\times10^{-5}\)
(b) \(2.7\times10^{-14}\)
(c) \(2.7\times10^{-10}\)
(d) \(9\times10^{-15}\)
Show Answer & Explanation

Answer: (b) \(2.7\times10^{-14}\)

Explanation:
1. Fraction \(=(r_N/r_A)^3 = (1.5\times10^{-15}/5\times10^{-11})^3 = (3\times10^{-5})^3 = 2.7\times10^{-14}\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 4: The radius of a nucleus of mass number 125 (take \(R_0 = 1.3\times10^{-15}\) m) is
(a) \(1.6\times10^{-14}\) m
(b) \(1.3\times10^{-14}\) m
(c) \(6.5\times10^{-15}\) m
(d) \(5.0\times10^{-15}\) m
Show Answer & Explanation

Answer: (c) \(6.5\times10^{-15}\) m

Explanation:
1. \(R = 1.3\times10^{-15}\times 125^{1/3} = 1.3\times10^{-15}\times5 = 6.5\times10^{-15}\) m.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 5: The mass to charge ratio (m/e) of a cation is \(1.5\times10^{-8}\) kg/C. If the cation carries one unit of charge, its mass is
(a) \(2.4\times10^{-19}\) g
(b) \(2.4\times10^{-24}\) g
(c) \(2.4\times10^{-27}\) g
(d) \(1.5\times10^{-8}\) g
Show Answer & Explanation

Answer: (b) \(2.4\times10^{-24}\) g

Explanation:
1. \(m = 1.5\times10^{-8}\times1.6\times10^{-19} = 2.4\times10^{-27}\) kg \(= 2.4\times10^{-24}\) g.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 6: Rutherford's scattering experiment with a gold foil showed that
(a) the positive charge and nearly all the mass are concentrated in a very small nucleus
(b) the atom is a uniform sphere of positive charge
(c) electrons are embedded in a positive sphere
(d) all the \(\alpha\)-particles are reflected back
Show Answer & Explanation

Answer: (a) the positive charge and nearly all the mass are concentrated in a very small nucleus

Explanation:
1. Most \(\alpha\)-particles passed straight through (empty space) and a few were scattered through large angles, which needs a tiny, massive, positively charged centre.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 7: An electromagnetic wave has a frequency of 400 MHz (as used in body scanners). Its wavelength is
(a) 0.75 cm
(b) 1.5 m
(c) 0.75 m
(d) 2 cm
Show Answer & Explanation

Answer: (c) 0.75 m

Explanation:
1. \(\lambda = c/\nu = 3\times10^{8}/4\times10^{8} = 0.75\) m.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 8: Radiation of wavelength 242 nm is just sufficient to ionise a sodium atom. The ionisation energy of sodium in kJ mol\(^{-1}\) is
(a) 247
(b) 494.65
(c) 400
(d) 600
Show Answer & Explanation

Answer: (b) 494.65

Explanation:
1. \(E = N_A hc/\lambda = 6.022\times10^{23}\times6.626\times10^{-34}\times3\times10^{8}/242\times10^{-9} = 4.9465\times10^{5}\) J/mol \(= 494.65\) kJ/mol.
2. Only option (b) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 9: Light of wavelength \(\lambda\) falls on a metal whose work function is \(hc/\lambda_0\). Photoelectric emission occurs only if
(a) \(\lambda \le \lambda_0\)
(b) \(\lambda \ge \lambda_0\)
(c) \(\lambda \ge 2\lambda_0\)
(d) \(\lambda \le \lambda_0/2\)
Show Answer & Explanation

Answer: (a) \(\lambda \le \lambda_0\)

Explanation:
1. Emission needs \(hc/\lambda \ge hc/\lambda_0\), which gives \(\lambda \le \lambda_0\).
2. Only option (a) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 10: Which of the following statements about the photoelectric effect is NOT correct?
(a) The number of ejected electrons depends on the intensity of the light
(b) The kinetic energy of ejected electrons depends on the intensity of the light
(c) The kinetic energy of ejected electrons depends on the frequency of the light
(d) There is a threshold frequency below which no emission occurs
Show Answer & Explanation

Answer: (b) The kinetic energy of ejected electrons depends on the intensity of the light

Explanation:
1. The maximum kinetic energy is \(h(\nu-\nu_0)\), which depends on frequency and not on intensity
2. Intensity only changes the number of electrons.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 11: The number of photons of radiation of frequency \(5\times10^{13}\) s\(^{-1}\) needed to melt 1 mole of ice (latent heat 330 J/g, \(h = 6.6\times10^{-34}\) J s) is
(a) \(1.8\times10^{23}\)
(b) \(1.8\times10^{22}\)
(c) \(1.8\times10^{24}\)
(d) \(5.4\times10^{23}\)
Show Answer & Explanation

Answer: (a) \(1.8\times10^{23}\)

Explanation:
1. Heat needed \(= 330\times18 = 5940\) J. Photon energy \(= 6.6\times10^{-34}\times5\times10^{13} = 3.3\times10^{-20}\) J. Number \(= 5940/3.3\times10^{-20} = 1.8\times10^{23} = 18\times10^{22}\).
2. Only option (a) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 12: The work function of a metal is 40 eV. The longest wavelength that can eject photoelectrons is nearest to
(a) 310 nm
(b) 12.4 nm
(c) 124 nm
(d) 31 nm
Show Answer & Explanation

Answer: (d) 31 nm

Explanation:
1. \(\lambda = 12400\ \text{eV\,\AA}/40\ \text{eV} = 310\) A = 31 nm.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 13: The energy of 100 photons of wavelength 2000 A is
(a) \(9.94\times10^{-19}\) J
(b) \(9.94\times10^{-15}\) J
(c) \(9.94\times10^{-17}\) J
(d) \(3.3\times10^{-17}\) J
Show Answer & Explanation

Answer: (c) \(9.94\times10^{-17}\) J

Explanation:
1. One photon: \(hc/\lambda = 6.626\times10^{-34}\times3\times10^{8}/2\times10^{-7} = 9.94\times10^{-19}\) J. For 100 photons, \(9.94\times10^{-17}\) J.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 14: Each Br\(_2\) molecule absorbs one photon of wavelength 5000 A to dissociate. The energy needed per mole is nearest to
(a) 239 kJ
(b) 120 kJ
(c) 478 kJ
(d) 24 kJ
Show Answer & Explanation

Answer: (a) 239 kJ

Explanation:
1. \(E = N_Ahc/\lambda = 6.022\times10^{23}\times6.626\times10^{-34}\times3\times10^{8}/5\times10^{-7} = 2.39\times10^{5}\) J = 239 kJ/mol.
2. Only option (a) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 15: Two bulbs emit red light (8000 A) and yellow light (4000 A) with the same number of photons per second. If the red bulb is rated 100 W, the yellow bulb is rated
(a) 50 W
(b) 100 W
(c) 400 W
(d) 200 W
Show Answer & Explanation

Answer: (d) 200 W

Explanation:
1. Power \(=\) photons per second \(\times\, hc/\lambda\), and photon energy is inversely proportional to wavelength, so the yellow bulb gives twice the power: 200 W.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 16: A light of frequency \(4\times10^{16}\) Hz ejects electrons with twice the maximum kinetic energy of those ejected by light of \(2.5\times10^{16}\) Hz from the same metal. The threshold frequency is
(a) \(1.5\times10^{16}\) Hz
(b) \(2\times10^{16}\) Hz
(c) \(1\times10^{16}\) Hz
(d) \(5\times10^{15}\) Hz
Show Answer & Explanation

Answer: (c) \(1\times10^{16}\) Hz

Explanation:
1. \(4\times10^{16}-\nu_0 = 2(2.5\times10^{16}-\nu_0)\) gives \(\nu_0 = 1\times10^{16}\) Hz.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 17: Light of frequency \(3.2\times10^{16}\) Hz ejects electrons with twice the maximum kinetic energy of electrons ejected by light of \(2.0\times10^{16}\) Hz from the same metal. The threshold frequency is
(a) \(2.4\times10^{16}\) Hz
(b) \(8.0\times10^{16}\) Hz
(c) \(7.2\times10^{16}\) Hz
(d) \(8.0\times10^{15}\) Hz
Show Answer & Explanation

Answer: (d) \(8.0\times10^{15}\) Hz

Explanation:
1. \(3.2\times10^{16}-\nu_0 = 2(2.0\times10^{16}-\nu_0)\) gives \(\nu_0 = 0.8\times10^{16} = 8.0\times10^{15}\) Hz.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 18: The work function of a metal is 3.1 eV (hc = 12400 eV A). Radiation of 300 nm strikes it. The maximum kinetic energy of the emitted electron is
(a) 4.13 eV
(b) 3.1 eV
(c) 1.03 eV
(d) 0.5 eV
Show Answer & Explanation

Answer: (c) 1.03 eV

Explanation:
1. Photon energy \(= 12400/3000 = 4.13\) eV. \(KE = 4.13-3.1 = 1.03\) eV. (The threshold wavelength is 4000 A.)
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 19: The threshold wavelength of sodium is 6500 A. When light of wavelength 360 A falls on it, the kinetic energy of the ejected electron is nearest to
(a) \(5.2\times10^{-18}\) erg
(b) \(5.2\times10^{-11}\) erg
(c) \(3.4\times10^{-11}\) erg
(d) \(5.2\times10^{-21}\) erg
Show Answer & Explanation

Answer: (b) \(5.2\times10^{-11}\) erg

Explanation:
1. \(KE = hc(1/360-1/6500)\ \text{\AA}^{-1} = 12400\,(2.778\times10^{-3}-1.54\times10^{-4}) = 32.5\) eV \(= 5.2\times10^{-18}\) J \(= 5.2\times10^{-11}\) erg.
2. Only option (b) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 20: Radiation from the 4 to 2 transition of hydrogen falls on a metal of work function 2.5 eV. The maximum kinetic energy of the photoelectrons is
(a) 2.55 eV
(b) 0.5 eV
(c) 5.05 eV
(d) 0.05 eV
Show Answer & Explanation

Answer: (d) 0.05 eV

Explanation:
1. Photon energy \(= 13.6(1/4-1/16) = 2.55\) eV, so \(KE = 2.55-2.5 = 0.05\) eV.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 21: A beam of white light falls on potassium after dispersion by a prism. The correct decreasing order of the maximum kinetic energy of the emitted electrons for different colours is
(a) red > orange > blue > violet
(b) blue > violet > green > yellow
(c) violet > blue > orange > red
(d) orange > yellow > blue > violet
Show Answer & Explanation

Answer: (c) violet > blue > orange > red

Explanation:
1. Maximum kinetic energy rises with frequency; the frequency order is violet > blue > green > yellow > orange > red.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 22: The ratio of the radii of the first Bohr orbits of H, He\(^+\), Li\(^{2+}\), Be\(^{3+}\) follows the order
(a) Be\(^{3+}\) > Li\(^{2+}\) > He\(^+\) > H
(b) H > He\(^+\) > Li\(^{2+}\) > Be\(^{3+}\)
(c) He\(^+\) > Be\(^{3+}\) > Li\(^{2+}\) > H
(d) He\(^+\) > H > Li\(^{2+}\) > Be\(^{3+}\)
Show Answer & Explanation

Answer: (b) H > He\(^+\) > Li\(^{2+}\) > Be\(^{3+}\)

Explanation:
1. \(r_1\propto1/Z\): Z = 1, 2, 3, 4, so the radius falls from H to Be\(^{3+}\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 23: A circular orbit of the hydrogen atom has a diameter of 20 nm. The orbit number is nearest to
(a) 10
(b) 12
(c) 14
(d) 16
Show Answer & Explanation

Answer: (c) 14

Explanation:
1. \(r = 100\) A \(= 0.529n^2\) A, so \(n^2 = 189\) and \(n \approx 13.7 \approx 14\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

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