Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

(c) According to wave mechanics, the ground state angular momentum is equal to $${h \over {2\pi }}$$.

(d) The plot of $$\psi $$ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

(c) According to wave mechanics, the ground state angular momentum is equal to $${h \over {2\pi }}$$.

(d) The plot of $$\psi $$ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

A

(a), (b)

B

(a), (d)

C

(b), (c)

D

(a), (c)

2

The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He^{+} ion in eV is :

A

$$-$$ 6.04

B

$$-$$ 54.4

C

$$-$$ 27.2

D

$$-$$ 3.4

(E)_{nth} = (E_{GND})_{H} . $${{{Z^2}} \over {{n^2}}}$$

E_{3rd} (He^{+}) = ($$-$$13.6eV) . $${{{2^2}} \over {{3^2}}}$$ = $$-$$ 6.04 eV

E

3

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose?

[R_{H} = 1 $$ \times $$ 10^{5} cm^{–1}, h = 6.6 $$ \times $$ 10^{–34} Js, c = 3 $$ \times $$ 10^{8} ms^{–1}]

[R

A

Balmer, $$\infty $$ $$ \to $$ 2

B

Paschen, 5 $$ \to $$ 3

C

Paschen, $$\infty $$ $$ \to $$ 3

D

Lyman, $$\infty $$ $$ \to $$ 1

Given, R_{H} = 1 $$ \times $$ 10^{5} cm^{–1}

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10^{-5} cm

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10^{-7} cm $$ \times $$ 100

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 100 nm

We know,

$${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$$

$$ \Rightarrow $$ $$\lambda $$ = $${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

[For H atom Z = 1]

$$ \Rightarrow $$ $$\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

$$ \Rightarrow $$ $$\lambda $$ = $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

Given, $$\lambda $$ = 900 nm

$$ \therefore $$ $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$ = 900

$$ \Rightarrow $$ $${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$

By checking each options you can see

when n_{L} = 3 and n_{H} = $$\infty $$ then

$${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$

$$ \therefore $$ Option C is correct.

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10

$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 100 nm

We know,

$${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$$

$$ \Rightarrow $$ $$\lambda $$ = $${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

[For H atom Z = 1]

$$ \Rightarrow $$ $$\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

$$ \Rightarrow $$ $$\lambda $$ = $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$

Given, $$\lambda $$ = 900 nm

$$ \therefore $$ $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$ = 900

$$ \Rightarrow $$ $${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$

By checking each options you can see

when n

$${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$

$$ \therefore $$ Option C is correct.

4

The de Broglie wavelength ($$\lambda $$) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v_{0} is threshold frequency] :

A

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{3 \over 2}}}}}$$

B

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 4}}}}}$$

C

$$\lambda \,\infty \,{1 \over {\left( {v - {v_0}} \right)}}$$

D

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$

By photoelectric effect

KE = h$$\gamma $$ - h$$\gamma $$_{o} ....(1)

de broglie wavelength,

$$\lambda $$ = $${h \over {mv}}$$ = $${h \over {\sqrt {2m \times K.E} }}$$ ...(2)

Using equation (1) and (2), we get

$$\lambda $$ = $${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$$

$$ \therefore $$ $$\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$

KE = h$$\gamma $$ - h$$\gamma $$

de broglie wavelength,

$$\lambda $$ = $${h \over {mv}}$$ = $${h \over {\sqrt {2m \times K.E} }}$$ ...(2)

Using equation (1) and (2), we get

$$\lambda $$ = $${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$$

$$ \therefore $$ $$\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$

Number in Brackets after Paper Name Indicates No of Questions

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Basics of Organic Chemistry *keyboard_arrow_right*

Hydrocarbons *keyboard_arrow_right*

Haloalkanes and Haloarenes *keyboard_arrow_right*

Alcohols, Phenols and Ethers *keyboard_arrow_right*

Aldehydes, Ketones and Carboxylic Acids *keyboard_arrow_right*

Compounds Containing Nitrogen *keyboard_arrow_right*

Polymers *keyboard_arrow_right*

Biomolecules *keyboard_arrow_right*

Chemistry in Everyday Life *keyboard_arrow_right*

Practical Organic Chemistry *keyboard_arrow_right*

Some Basic Concepts of Chemistry *keyboard_arrow_right*

Structure of Atom *keyboard_arrow_right*

Gaseous State *keyboard_arrow_right*

Colloidal State *keyboard_arrow_right*

Redox Reactions *keyboard_arrow_right*

Thermodynamics *keyboard_arrow_right*

Equilibrium *keyboard_arrow_right*

Solid State & Surface Chemistry *keyboard_arrow_right*

Solutions *keyboard_arrow_right*

Electrochemistry *keyboard_arrow_right*

Chemical Kinetics and Nuclear Chemistry *keyboard_arrow_right*

Periodic Table & Periodicity *keyboard_arrow_right*

Chemical Bonding & Molecular Structure *keyboard_arrow_right*

s-Block Elements *keyboard_arrow_right*

Isolation of Elements *keyboard_arrow_right*

Hydrogen *keyboard_arrow_right*

p-Block Elements *keyboard_arrow_right*

d and f Block Elements *keyboard_arrow_right*

Coordination Compounds *keyboard_arrow_right*

Environmental Chemistry *keyboard_arrow_right*