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Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

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.

2

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

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

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

3

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

4

If the de Broglie wavelength of the electron in n^{th} Bohr orbit in a hydrogenic atom is equal to 1.5 $$\pi $$a_{0} (a_{0} is Bohr radius), then the value of n/z is -

A

0.75

B

0.40

C

1.50

D

1.0

According to debroglie hypothesis,

2$$\pi $$r_{n} = *n*$$\lambda $$

$$ \therefore $$ $$\lambda $$ = $${{2\pi {r_n}} \over n}$$

According to the question,

$${{2\pi {r_n}} \over n}$$ = 1.5 $$\pi $$*a*_{0}

We know,

*r*_{n} = 0.529 $${{{n^2}} \over Z}$$

=*a*_{0} $$ \times $$ $${{{n^2}} \over Z}$$

$$ \therefore $$ $${{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}$$ = 1.5 $$\pi $$*a*_{0}

$$ \Rightarrow $$ $${n \over Z}$$ = 0.75

2$$\pi $$r

$$ \therefore $$ $$\lambda $$ = $${{2\pi {r_n}} \over n}$$

According to the question,

$${{2\pi {r_n}} \over n}$$ = 1.5 $$\pi $$

We know,

=

$$ \therefore $$ $${{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}$$ = 1.5 $$\pi $$

$$ \Rightarrow $$ $${n \over Z}$$ = 0.75

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