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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2009

MCQ (Single Correct Answer)
The surface of a metal is illuminated with the light of $$400$$ $$nm.$$ The kinetic energy of the ejected photoelectrons was found to be $$1.68$$ $$eV.$$ The work function of the metal is : $$\left( {hc = 1240eV.nm} \right)$$
A
$$1.41$$ $$eV$$
B
$$1.51$$ $$eV$$
C
$$1.68$$ $$eV$$
D
$$3.09$$ $$eV$$

Explanation

$$\lambda = 400nm,\,\,hc = 1240eV.nm,\,\,K.E. = 1.68\,eV$$

We know that $${{hc} \over \lambda } - W = K.E \Rightarrow W = {{hc} \over \lambda } - K.E$$

$$ \Rightarrow W = {{1240} \over {400}} - 1.68$$

$$ = 3.1 - 1.68 = 1.42\,eV$$
2

AIEEE 2008

MCQ (Single Correct Answer)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

If a strong diffraction peak is observed when electrons are incident at an angle $$'i'$$ from the normal to the crystal planes with distance $$'d'$$ between them (see figure), de Broglie wavelength $${\lambda _{dB}}$$ of electrons can be calculated by the relationship ($$n$$ is an integer)

A
$$d\,\sin \,i = n{\lambda _{dB}}$$
B
$$2d\,\cos \,i = n{\lambda _{dB}}$$
C
$$2d\,\sin \,i = n{\lambda _{dB}}$$
D
$$d\,\cos \,i = n{\lambda _{dB}}$$

Explanation

$$2d\,\cos \,i = n{\lambda _{dB}}$$
3

AIEEE 2008

MCQ (Single Correct Answer)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Electrons accelerated by potential $$V$$ are diffracted from a crystal. If $$d = 1\mathop A\limits^ \circ $$ and $$i = {30^ \circ },\,\,\,V$$ should be about
$$\left( {h = 6.6 \times {{10}^{ - 34}}Js,{m_e} = 9.1 \times {{10}^{ - 31}}kg,\,e = 1.6 \times {{10}^{ - 19}}C} \right)$$

A
$$2000$$ $$V$$
B
$$50$$ $$V$$
C
$$500$$ $$V$$
D
$$1000$$ $$V$$

Explanation

Using Bragg's equation $$2d$$ $$\sin \theta = n\lambda $$

Here $$n=1,$$ $$\theta = 90 - i = 90 - 30 = {60^ \circ }$$

$$\therefore$$ $$2d$$ $$\sin \,\theta \, = \lambda \,\,\,\,\,\,\,\,\,\,....\left( i \right)$$



Also, $$\lambda = {{12.27} \over {\sqrt V }} \times {10^{ - 10}}m\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(i)$$ & $$(ii)$$ $$2 \times {10^{ - 10}} \times \sin {60^ \circ } = {{12.27} \over {\sqrt V }} \times {10^{ - 10}}$$

$$V = {{{{\left( {12.27} \right)}^2}} \over 3} = 50V$$
4

AIEEE 2008

MCQ (Single Correct Answer)
In an experiment, electrons are made to pass through a narrow slit of width $$'d'$$ comparable to their de Broglie wavelength. They are detected on a screen at a distance $$'D'$$ from the slit (see figure).

Which of the following graphs can be expected to represent the number of electrons $$'N'$$ detected as a function of the detector position $$'y'\left( {y = 0} \right.$$ corresponds to the middle of the slit$$\left. \, \right)$$

A
B
C
D

Explanation

The electron beam will be diffracted and the maxima is obtained at $$y=0.$$ Also the distance between the first minima on both side will be greater than $$d.$$

Questions Asked from Dual Nature of Radiation

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