Joint Entrance Examination

Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

Surface of certain metal is first illuminated with light of wavelength $$\lambda $$_{1} = 350 nm and then, by light of wavelength $$\lambda $$_{2} = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

(Energy of photon n = $${{1240} \over {\lambda (in\,mm)}}$$eV)

(Energy of photon n = $${{1240} \over {\lambda (in\,mm)}}$$eV)

A

1.8

B

2.5

C

5.6

D

1.4

Let speed of photon electron in first case is 2v

then in the second case speed is v.

For first case

$${{hc} \over {{\lambda _1}}} = \phi + {1 \over 2}$$m(2v)^{2}

For second case,

$${{hc} \over {{\lambda _2}}} = \phi + {1 \over 2}$$mv^{2}

$$ \therefore $$ $${{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m \times 4{v^2}} \over {{1 \over 2}m{v^2}}}$$

$$ \Rightarrow $$ $${{{hc} \over {{\lambda _1}}} - \phi }$$ = 4$$\left( {{{hc} \over {{\lambda _2}}} - \phi } \right)$$

$$ \Rightarrow $$ $${{4hc} \over {{\lambda _2}}}$$ $$-$$ $${{hc} \over {{\lambda _1}}}$$ = 3$$\phi $$

$$ \Rightarrow $$ $$\phi $$ $$=$$ $${{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right)$$

$$ \Rightarrow $$ $$\phi $$ $$=$$ $${{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} \right)$$

$$ \Rightarrow $$ $$\phi $$ = 1.8 eV

then in the second case speed is v.

For first case

$${{hc} \over {{\lambda _1}}} = \phi + {1 \over 2}$$m(2v)

For second case,

$${{hc} \over {{\lambda _2}}} = \phi + {1 \over 2}$$mv

$$ \therefore $$ $${{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m \times 4{v^2}} \over {{1 \over 2}m{v^2}}}$$

$$ \Rightarrow $$ $${{{hc} \over {{\lambda _1}}} - \phi }$$ = 4$$\left( {{{hc} \over {{\lambda _2}}} - \phi } \right)$$

$$ \Rightarrow $$ $${{4hc} \over {{\lambda _2}}}$$ $$-$$ $${{hc} \over {{\lambda _1}}}$$ = 3$$\phi $$

$$ \Rightarrow $$ $$\phi $$ $$=$$ $${{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right)$$

$$ \Rightarrow $$ $$\phi $$ $$=$$ $${{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} \right)$$

$$ \Rightarrow $$ $$\phi $$ = 1.8 eV

2

The magnetic field associated with a light wave is given, at the origin, by B = B_{0} [sin(3.14 $$ \times $$ 10^{7})ct + sin(6.28 $$ \times $$ 10^{7})ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?

(c = 3 $$ \times $$ 10^{8} ms^{$$-$$1}, h = 6.6 $$ \times $$ 10^{$$-$$34}J-s)

(c = 3 $$ \times $$ 10

A

6.82 eV

B

12.5 eV

C

8.52 eV

D

7.72 eV

Given that,

B = B_{0}[sin (3.14 $$ \times $$ 10^{7}) ct + sin(6.28 $$ \times $$ 10^{7}) ct]

This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ($$\phi $$) = 4.7 eV

Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As,

K_{max} = E_{max} $$-$$ $$\phi $$

Emax = hF

= h $$ \times $$ $${\omega \over {2\pi }}$$

= 6.6 $$ \times $$ 10^{$$-$$34} $$ \times $$ $${{6.28 \times {{10}^7} \times 3 \times {{10}^8}} \over {2\pi }}$$

= 6.6$$ \times $$ 3 $$ \times $$ 10^{$$-$$19} J

= $${{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV$$

= 12.375 $$eV$$

$$ \therefore $$ K_{max} = 12.375 $$-$$ 4.7

= 7.675 $$eV$$

$$ \simeq $$ 7.7 $$eV$$

B = B

This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ($$\phi $$) = 4.7 eV

Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As,

K

Emax = hF

= h $$ \times $$ $${\omega \over {2\pi }}$$

= 6.6 $$ \times $$ 10

= 6.6$$ \times $$ 3 $$ \times $$ 10

= $${{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV$$

= 12.375 $$eV$$

$$ \therefore $$ K

= 7.675 $$eV$$

$$ \simeq $$ 7.7 $$eV$$

3

A metal plate of area 1 $$ \times $$ 10^{–4} m^{2} is illuminated by a radiation of intensity 16 mW/m^{2}. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons.
The number of emitted photoelectrons per second and their maximum energy, respectively, will be

A

10^{14} and 10 eV

B

10^{12} and 5 eV

C

10^{11} and 5 eV

D

10^{10} and 5 eV

Given, $$\phi $$ = 5 eV, E = 10 eV

$$ \therefore $$ KE = E – $$\phi $$ = 10 – 5 = 5 eV

Energy incident on the plate,

E' = $$\left( {16 \times {{10}^{ - 3}}} \right)\left( {1 \times {{10}^{ - 4}}} \right)$$

= 16 $$ \times $$ 10^{-7} J/s

= 10^{13} eV/s

Number of photons emitted per second

= 10% of $${{E'} \over E} = 0.1 \times {{{{10}^{13}}} \over {10}}$$ = 10^{11} s^{-1}

$$ \therefore $$ KE = E – $$\phi $$ = 10 – 5 = 5 eV

Energy incident on the plate,

E' = $$\left( {16 \times {{10}^{ - 3}}} \right)\left( {1 \times {{10}^{ - 4}}} \right)$$

= 16 $$ \times $$ 10

= 10

Number of photons emitted per second

= 10% of $${{E'} \over E} = 0.1 \times {{{{10}^{13}}} \over {10}}$$ = 10

4

If the de Broglie wavelength of an electron is equal to the 10^{–3} times the wavelength of a photon of frequency 6 $$ \times $$ 10^{14} Hz, then the speed of electron is equal to : (Speed of light = 3 $$ \times $$ 10^{8} m/s, Planck's constant = 6.63 $$ \times $$
10^{–34} J.s, Mass of electron = 9.1 $$ \times $$ 10^{–31} kg)

A

1.7 $$ \times $$ 10^{6} m/s

B

1.45 $$ \times $$ 10^{6} m/s

C

1.1 $$ \times $$ 10^{6} m/s

D

1.8 $$ \times $$ 10^{6} m/s

$${h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)$$

v $$ = {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$$

v $$ = 1.45 \times {10^6}$$ m/s

v $$ = {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$$

v $$ = 1.45 \times {10^6}$$ m/s

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