Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s^{$$-$$1}) of He atom after the photon absorption is __________.

(Assume : Momentum is conserved when photon is absorbed.

Use : Planck constant = 6.6 $$\times$$ 10^{$$-$$34} J s, Avogadro number = 6 $$\times$$ 10^{23} mol^{$$-$$1}, Molar mass of He = 4 g mol^{$$-$$1})

(Assume : Momentum is conserved when photon is absorbed.

Use : Planck constant = 6.6 $$\times$$ 10

Your Input ________

Correct Answer is **30**

Wavelength of photon absorbed, $$\lambda$$ = 330 nm = 330 $$\times$$ 10^{$$-$$9} m

Planck's constant, h = 6.6 $$\times$$ 10^{$$-$$34} J s

Molar mass of He, M = 4 g mol^{$$-$$1} = 4 $$\times$$ 10^{$$-$$3} kg mol^{$$-$$1}

Avogadro number, N_{A} = 6 $$\times$$ 10^{23} mol^{$$-$$1}

Mass of one atom of He, $$m = {M \over {{N_A}}}$$

$$ = {{4 \times {{10}^{ - 3}}} \over {6 \times {{10}^{23}}}} = {2 \over 3} \times {10^{ - 26}}$$ kg

Velocity, = V cm/s.

Using de-Broglie equation,

$$\lambda = {h \over {mv}}$$

$$ \Rightarrow v = {h \over {m\lambda }} = {{6.6 \times {{10}^{ - 34}}} \over {2/3 \times {{10}^{ - 26}} \times 330 \times {{10}^{ - 9}}}}$$

$$ = {{6.6 \times 3 \times {{10}^{ - 34}} \times {{10}^{35}}} \over {2 \times 330}} = 0.03 \times 10 = 0.3$$ m/s

= 30 cm/s

Planck's constant, h = 6.6 $$\times$$ 10

Molar mass of He, M = 4 g mol

Avogadro number, N

Mass of one atom of He, $$m = {M \over {{N_A}}}$$

$$ = {{4 \times {{10}^{ - 3}}} \over {6 \times {{10}^{23}}}} = {2 \over 3} \times {10^{ - 26}}$$ kg

Velocity, = V cm/s.

Using de-Broglie equation,

$$\lambda = {h \over {mv}}$$

$$ \Rightarrow v = {h \over {m\lambda }} = {{6.6 \times {{10}^{ - 34}}} \over {2/3 \times {{10}^{ - 26}} \times 330 \times {{10}^{ - 9}}}}$$

$$ = {{6.6 \times 3 \times {{10}^{ - 34}} \times {{10}^{35}}} \over {2 \times 330}} = 0.03 \times 10 = 0.3$$ m/s

= 30 cm/s

2

Numerical

The figure below is the plot of potential energy versus internuclear distance (d) of H_{2} molecule in the electronic ground state. What is the value of the net potential energy E_{0} (as indicated in the figure) in kJ mol^{-1}, for d = d_{0} at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of H atom is taken as zero when its electron and the nucleus are infinitely far apart.

Use Avogadro constant as 6.023 $$ \times $$ 10^{23} mol^{-1}.

Use Avogadro constant as 6.023 $$ \times $$ 10

Your Input ________

Correct Answer is **$$-$$5242.41**

Given that, electrons and nucleus are at infinite distance, so potential energy of H-atom is taken as zero.

Therefore, according to Bohr's model, potential energy of a H-atom with electron in its ground state = $$-$$27.2 eV

At d = d_{0}, nucleus-nucleus and electron-electron repulsion is absent.

Hence, potential energy will be calculated for 2 H atoms = $$-$$2 $$ \times $$ 27.2 eV = $$-$$54.4 eV

Potential energy of 1 mol H atoms in kJ

= $${{54.4 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}$$

= $$ - 5242.4192$$ kJ/mol

Therefore, according to Bohr's model, potential energy of a H-atom with electron in its ground state = $$-$$27.2 eV

At d = d

Hence, potential energy will be calculated for 2 H atoms = $$-$$2 $$ \times $$ 27.2 eV = $$-$$54.4 eV

Potential energy of 1 mol H atoms in kJ

= $${{54.4 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}$$

= $$ - 5242.4192$$ kJ/mol

3

Numerical

Not considering the electronic spin, the degeneracy of the second excited state( n = 3) of H atom is 9, while
the degeneracy of the second excited state of H^{–} is

Your Input ________

Correct Answer is **3**

4

Numerical

In an atom, the total number of electrons having quantum numbers n = 4, |m_{l}| = 1 and m_{s} = –1/2 is

Your Input ________

Correct Answer is **6**

As n = 4, so l = 0, 1, 2, 3 which implies the orbitals are 4s, 4p, 4d and 4f.

Now, |m_{l}
| = 1 implies m_{l}
= +1 and −1. Therefore, l can be
3, 2, 1, as for l = 0, m_{l}
= 0.

For l = 1, m_{l}
= −1, 0, +1

For l = 2, m_{l}
= −2, −1, 0, +1, +2

For l = 3, m_{l}
= −3, −2, −1, 0, +1, +2, +3

On those following papers in Numerical

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2021 Paper 2 Online (1)

JEE Advanced 2020 Paper 2 Offline (1)

JEE Advanced 2015 Paper 1 Offline (1)

JEE Advanced 2014 Paper 1 Offline (1)

JEE Advanced 2013 Paper 1 Offline (1)

IIT-JEE 2011 Paper 1 Offline (2)

IIT-JEE 1994 (1)

IIT-JEE 1985 (1)

Some Basic Concepts of Chemistry

Structure of Atom

States of Matter

Thermodynamics

Equilibrium

Solid State & Surface Chemistry

Solutions

Electrochemistry

Chemical Kinetics and Nuclear Chemistry

Gaseous State

Redox Reactions

Periodic Table & Periodicity

Chemical Bonding & Molecular Structure

s-Block Elements

Isolation of Elements

Hydrogen

p-Block Elements

d and f Block Elements

Coordination Compounds

Salt Analysis

Basics of Organic Chemistry

Hydrocarbons

Haloalkanes and Haloarenes

Alcohols, Phenols and Ethers

Aldehydes, Ketones and Carboxylic Acids

Compounds Containing Nitrogen

Polymers

Biomolecules

Chemistry in Everyday Life