 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2010

If a source of power $4kW$ produces ${10^{20}}$ photons/second, the radiation belongs to a part of the spectrum called
A
$X$ -rays
B
ultraviolet rays
C
microwaves
D
$\gamma$ - rays

Explanation

Power, $P = {{nhv} \over t}$

$\Rightarrow v = {{P \times t} \over {nh}}$

$= {{4 \times {{10}^3} \times 1} \over {{{10}^{20}} \times 6.63 \times {{10}^{ - 34}}}} = 6 \times {10^{16}}Hz$
2

AIEEE 2010

Statement - $1$ : When ultraviolet light is incident on a photocell, its stopping potential is ${V_0}$ and the maximum kinetic energy of the photoelectrons is ${K_{\max }}$. When the ultraviolet light is replaced by $X$-rays, both ${V_0}$ and ${K_{\max }}$ increase.

Statement - $2$ : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

A
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is the correct explanation of Statement - $1$
B
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is not the correct explanation of Statement - $1$
C
Statement - $1$ is is false, Statement - $2$ is true
D
Statement - $1$ is is true, Statement - $2$ is false

Explanation

we know that

$e{V_0} = {K_{\max }} = hv - \phi$

where, $\phi$ is the work function.

Hence, as v increases (note that frequency of $X$-rays is greater than that of $U.V.$ rays), both ${V_0}$ and ${K_{\max }}$ increase. So statement - $1$ is correct
3

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

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

$= 3.1 - 1.68 = 1.42\,eV$
4

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

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