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

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

2

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.

3

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

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

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

4

What is the work function of the metal if the light of wavelength 4000$$\mathop A\limits^ \circ $$ generates photoelectrons of velocity 6 $$ \times $$ 10^{5} ms^{–1} from it ?

(Mass of electron = 9 $$ \times $$ 10^{–31} kg;

Velocity of light = 3 $$ \times $$ 10^{8} ms^{$$-$$1}

Plank's constant = 6.626 $$ \times $$ 10^{–34} Js;

Charge of electron = 1.6 $$ \times $$10^{–19} JeV^{–1})

(Mass of electron = 9 $$ \times $$ 10

Velocity of light = 3 $$ \times $$ 10

Plank's constant = 6.626 $$ \times $$ 10

Charge of electron = 1.6 $$ \times $$10

A

4.0 eV

B

0.9 eV

C

2.1 eV

D

3.1 eV

E = $$\phi $$ + K.E

$$ \Rightarrow $$ h$$\nu $$ = $$\phi $$ + $${1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $$\phi $$ = $$h\nu - {1 \over 2}m{v^2}$$

= $${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$$

= 3.35 $$ \times $$ 10^{-19} J

$$ \Rightarrow $$ $$\phi $$ = $${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV

= 2.0934 eV $$ \simeq $$ 2.1 eV

$$ \Rightarrow $$ h$$\nu $$ = $$\phi $$ + $${1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $$\phi $$ = $$h\nu - {1 \over 2}m{v^2}$$

= $${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$$

= 3.35 $$ \times $$ 10

$$ \Rightarrow $$ $$\phi $$ = $${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV

= 2.0934 eV $$ \simeq $$ 2.1 eV

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Chemistry in Everyday Life *keyboard_arrow_right*

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