Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $${{{\lambda _A}} \over {{\lambda _B}}}$$ is close to :

A

4.47

B

10.00

C

14.14

D

0.07

K.E. acquired by charge = K = qV

$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2mK} }}$$ = $${h \over {\sqrt {2mqV} }}$$

$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50} $$

$$ = 2 \times 7.07 = 14.14$$

$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2mK} }}$$ = $${h \over {\sqrt {2mqV} }}$$

$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50} $$

$$ = 2 \times 7.07 = 14.14$$

2

MCQ (Single Correct Answer)

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: ($${{{hc} \over e}}$$ = 1240 nm$$-$$V)

A

0.5 V

B

1.0 V

C

2.0 V

D

1.5 V

$${{hc} \over {{\lambda _1}}} = \phi + e$$V_{1} . . . (i)

$${{hc} \over {{\lambda _2}}} = \phi + e$$V_{2} . . . (ii)

(i) $$-$$ (ii)

hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$$ = e(V_{1} $$-$$ V_{2})

$$ \Rightarrow $$ V_{1} $$-$$ V_{2} = $${{hc} \over e}$$$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)$$

= (1240nm $$-$$ V) $${{100nm} \over {300nm \times 400nm}}$$

= 1V

$${{hc} \over {{\lambda _2}}} = \phi + e$$V

(i) $$-$$ (ii)

hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$$ = e(V

$$ \Rightarrow $$ V

= (1240nm $$-$$ V) $${{100nm} \over {300nm \times 400nm}}$$

= 1V

3

MCQ (Single Correct Answer)

If the de Broglie wavelength of an electron is equal to the 10^{â€“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^{â€“34} J.s, Mass of electron = 9.1 $$ \times $$ 10^{â€“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

4

MCQ (Single Correct Answer)

A metal plate of area 1 $$ \times $$ 10^{â€“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 â€“ $$\phi $$ = 10 â€“ 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 â€“ $$\phi $$ = 10 â€“ 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

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