Download our App

JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
(100k+ download)
1

AIEEE 2006

MCQ (Single Correct Answer)
The threshold frequency for a metallic surface corresponds to an energy of $6.2$ $eV$ and the stopping potential for a radiation incident on this surface is $5V.$ The incident radiation lies in
A
ultra-violet region
B
infra-red region
C
visible region
D
$x$-ray region

Explanation

$\phi = 6.2\,eV = 6.2 \times 1.6 \times {10^{ - 19}}J$

$V = 5\,\,volt,\,\,{{hc} \over \lambda } - \phi = e{V_0}$

$\Rightarrow \lambda = {{hc} \over {\phi + e{V_0}}}$

$= {{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6 \times {{10}^{ - 19}}\left( {6.2 + 5} \right)}} \approx {10^{ - 7}}\,m$

This range lies in ultra violet range.
2

AIEEE 2005

MCQ (Single Correct Answer)
A photocell is illuminated by a small bright source placed $1$ $m$ away. When the same source of light is placed ${1 \over 2}$ $m$ away, the number of electrons emitted by photo-cathode would
A
increases by a factor of $4$
B
decreases by a factor of $4$
C
increases by a factor of $2$
D
decreases by a factor of $2$

Explanation

$I \propto {1 \over {{r^2}}};{{{I_1}} \over {{I_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^2} = {1 \over 4}$

${I_2} \to 4\,\,$ times ${I_1}$

When intensity becomes 4 times, no. of photoelectrons emitted would increase by $4$ times, since number of electrons emitted per second is directly proportional to intensity.
3

AIEEE 2005

MCQ (Single Correct Answer)
If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor
A
$2$
B
${1 \over 2}$
C
${\sqrt 2 }$
D
${1 \over {\sqrt 2 }}$

Explanation

de-Broglie wavelength,

$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$

$\therefore$ $\lambda \propto {1 \over {\sqrt {K.E} }}$

If $K.E$ is doubled, wavelength becomes ${\lambda \over {\sqrt 2 }}$
4

AIEEE 2004

MCQ (Single Correct Answer)
According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $Vs$ the frequency, of the incident radiation gives as straight the whose slope
A
depends both on the intensity of the radiation and the metal used
B
depends on the intensity of the radiation
C
depends on the nature of the metal used
D
is the same for the all metals and independent of the intensity of the radiation

Explanation

From Equation $K.E = hv - \phi$

slope of graph of $K.E$ and $v$ is = $h$ (Plank's constant) which is same for all metals.

Joint Entrance Examination

JEE Advanced JEE Main

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET

Class 12