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JEE Main Test Series Atomic Structure: Previous Year Questions and Quick Formulas for Revision
Results: 10 Mar `21
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Sonal Vaid

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In JEE main examination, Atomic Structure is an important topic which further contains many sub-topics like an atomic model, atomic structure, Thomson Atomic Model, Dalton’s Atomic theory, Bohr’s Atomic theory, Rutherford Atomic theory, etc. the candidates appearing for the examination should be well prepared and should know JEE Main Chemistry Syllabus in detail. 

The atomic structure in JEE Main refers to the laws of its nucleus and the grouping of the electrons around it. Generally, 3-5 questions are asked from the atomic structure which have a weightage of about 10%-15%. Candidates should also go through JEE Main previous year’s question papers and should practice the papers to score well in the specific topic. 

Check JEE Main Study Notes on Atomic Structure

JEE Main Test Series Atomic Structure: Previously Asked Questions

Question 1- If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

Options-

(a) Two times

(b) Half

(c) One fourth

(d) Four times

Solution and explanation 

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.

Answer- (B) Half

Question 2 - Which of the following atoms and ions are isoelectronic i.e. have the same number of electrons as the neon atom 

Options-

A) F−

B) Oxygen atom

C) Mg

 D) N−

Solution and explanation

(F− )have the same number of electrons as the neon atom.

Answer: Option A (F-)

Question 3- The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol-1. The longest wavelength of light capable of breaking a single Cl–Cl bond is (C = 3×108 m/s and NA = 6.02×1023 mol-1)

Options-

(1) 494 nm

(2) 594 nm

(3) 640 nm

(4) 700 nm

Solution and explanation

We have B.E = 242KJ/Mol

E = hcNA/λ

∴ λ = hcNA/E

= 3×108×6.626×10-34×6.02×1023/(242×103)

= 0.494×10-3×103

= 494 nm

Answer- (A) 494 nm

Questions 4- Atoms consist of protons, neutrons, and electrons. If the mass of neutrons and electrons were made half and two times respectively to their actual masses, then the atomic mass of 6 c12

Options-

A) Will remain approximately the same

B) Will become approximately two times

C) Will remain approximately half

D) Will be reduced by 25%

Solution and explanation

No change by doubling the mass of electrons however by reducing the mass of neutron to half total atomic mass becomes 6+3 instead of 6+6. Thus reduced by 25%.

Answer: D Will be reduced by 25%

Question 5- Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)

Options-

(A) 2.5 nm

(B) 14.0 nm

(C) 0.032 nm

(D) 0.40 nm

Solution and explanation-

Given mp = 1.67×10-27kg

h = 6.63×10-34Js

v = 1.0×103ms-1

We know wavelength λ = h/mv

∴λ = 6.63×10-34/(1.67×10-27 × 1.0×103)

= 3.97×10-10

≈ 0.04nm

Answer- (D) 0.40 nm

Question 6- The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following

Options-

(A) n = 3 to n = 1

(B) n = 2 to n = 1

(C) n = 3 to n = 2

(D) n = 4 to n = 3

Solution and explanation

E = 13.6×4[(¼)-(1/16)]

= 10.2

E = hν

ν = 10.2/h

E = 13.6(1)[(1/n12-1/n22)]

10.2 = 13.6[(1/n12-1/n22)]

102/136 = (n22-n12)/n12n22

Substitute the given options and find n1 and n2

51/68 = (n22-n12)/n12n22

0.75 = (4-1)4 = ¾ = 0.75

Answer- (B) n = 2 to n = 1

Question 7- Discovery of the nucleus of an atom was due to the experiment carried out by 

Options-

A) Bohr

B) Mosley

C) Rutherford

D) Thomson

Solution and explanation-

Rutherford discovered nucleus.

Answer: C, Rutherford

Question 8- Which of the following sets of quantum numbers is correct for an electron present in 4f orbital?

Options-

(A) n = 4, l = 3, m = +4, s = +½

(B) n = 3, l = 2, m = -2, s = +½

(C) n = 4, l = 3, m = +1, s = +½

(D) n = 4, l = 4, m = -4, s = -½

Solution and explanation-

For a 4f orbital, n = 4 and l = 3.

Values of m = -3, -2, -1, 0, +1, +2, +3

ANswer- (C) n = 4, l = 3, m = +1, s = +½

Question 9- Which of the following statements in relation to the hydrogen atom is correct?

Options-

(A)3s orbital is lower in energy than 3p orbital

(B)3p orbital is lower in energy than 3d orbital

(C)3s and 3p orbitals are of lower energy than 3d orbital

(D)3s, 3p, and 3d orbitals all have the same energy

Solution and explanation-

The Auf-bau principle is not applicable for the Hydrogen atom.

Answer- (D) 3s, 3p and 3d orbitals all have the same energy

Question 10- The number of d-electrons retained in Fe2+ (At.no. of Fe = 26) ion is

Options-

(A) 4

(B) 5

(C) 6

(D) 3

Solution and explanation

Configuration of Fe2+ = 3d6 4s0

Answer- (C) 6

Question 11- Which of the following sets of quantum numbers represents the highest energy of an atom?

Options

(A)n=3, l =2, m=l, s= +½

(B)n=3, l =2, m=l, s= +½

(C)n=4, l =0, m=0, s= +½

(D)n=3, l =0, m=0, s= +½

Solution and explanation

The maximum value of (n +l) represents the high­est energy of the orbital.

Answer- (B) n=3, l =2, m=l, s= +½

Question 12- In the Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen

Options-

(A) 5 → 2

(B) 4 → 1

(C) 2→ 5

(D) 3→ 2

Solution and explanation-

The lines falling in the visible spectrum include the Balmer series. So the third line would be n1 = 2 and n2 = 5. Thus the transition is 5 → 2

Answer- (A) 5 → 2

Question 13- The series limit for Balmer series of H-spectra is

Options-

A) 3800

B) 4200

C) 3646

D) 4000

Solution and explanation-

The last line in any series is called the series limit. The Series limit for the Balmer series is 

Undefined control sequence \AA

Answer: C 3646

Question 14- As an electron moves away from the nucleus, its potential energy

Options-

A) Increases

B) Decreases

C) Remains constant

D) None of these

Answer: A Increases

Question 15- The third line in the Balmer series corresponds to an electronic transition between which Bohr’s orbits in hydrogen 

Options-

A) 5 ® 3

B) 5 ® 2

C) 4 ® 3

D) 4 ® 2

Solution and explanation 

In Balmer series of hydrogen atomic spectrum which electronic transition causes third line O→L, n2=5→n1=2

Answer: (B) 5 ® 2

Check JEE Main Sample Papers

JEE Main Atomic Structure: Quick Formulas

Here is a list of formulas that are very important for the students to know who is appearing for the JEE Main Atomic Structure examination.

Atomic structure formula:

1. The velocity of the electron in nth Bohr orbit: v = 2.18 \times × 10^{8} \frac{Z}{n} cm/sec.

2. The radius of nth Bohr orbital: r_{n}= 0.529 \frac{n^{2}}{z}A^{0}

3. The total energy of an electron in nth orbit: E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV

4. The kinetic energy of electron: -(total energy of electron): 13.6\frac{Z^{2}}{n^{2}}\ eV = -P.E/2

5.The potential energy of the electron: -27.2\frac{Z^{2}}{n^{2}}\ eV

6. Line Spectrum of Hydrogen-like atoms 

\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where R is called Rydberg constant, R = 1.097\ast10^{7} where Z= atomic number.

7. De-Broglie wavelength

\lambda = \frac{h}{mv}= \frac{h}{p}

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