### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2009

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

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

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

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 $d.$