1

### JEE Main 2016 (Online) 9th April Morning Slot

When photons of wavelength ${\lambda _1}$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength ${\lambda _2}$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength ${\lambda _1}$ is used then find the stopping potential for this case :
A
${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} - {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$
B
${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$
C
${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {3 \over {2{\lambda _1}}}} \right]$
D
${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$

## Explanation

We know,

Einstein's photoelectric equation,

$eV = {{hc} \over \lambda } - {\phi _0}$

and ${\phi _0}$, ${{hc} \over {{\lambda _0}}}$, where ${{\lambda _0}}$ is the threashhold wavelength.

$\therefore$   In first case,

eV $= {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}$      . . .(1)

and in second case,

3 eV $= {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$       . . .(2)

Now, let slopping potential = V1 when light of wavelength $\lambda$3 is used then,

eV1 $= {{hc} \over {{\lambda _3}}} - {{hc} \over {{\lambda _0}}}$       . . .(3)

From (1) and (2) get,

$3\left( {{{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}} \right) = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$

$\Rightarrow$   ${{3hc} \over {{\lambda _1}}}$ $-$ ${{hc} \over {{\lambda _2}}}$ $=$ ${{2hc} \over {{\lambda _0}}}$

$\Rightarrow$   ${{hc} \over {{\lambda _0}}}$ $=$ ${{3hc} \over {2{\lambda _1}}}$ $-$ ${{hc} \over {2{\lambda _2}}}$

Putting this value of ${{hc} \over {{\lambda _0}}}$ in equation 3,

eV1 $= {{hc} \over {{\lambda _3}}} - {{3hc} \over {2{\lambda _1}}} + {{hc} \over {2{\lambda _2}}}$

$\Rightarrow$   V1 $= {{hc} \over e}$ $\left[ {{1 \over {{\lambda _3}}} - {3 \over {2{\lambda _1}}} + {1 \over {2{\lambda _2}}}} \right]$
2

### JEE Main 2016 (Online) 10th April Morning Slot

A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and ${\lambda \over 2}.$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
A
${{hc} \over {3\lambda }}$
B
${{hc} \over {2\lambda }}$
C
${{hc} \over {\lambda }}$
D
${3\,{hc} \over {\lambda }}$

## Explanation

We know,

Einstein's photo electric equation,

(KE)max = ${{hc} \over \lambda }$ $-$ $\phi$0

In first case,

K = ${{hc} \over \lambda }$ $-$ $\phi$0           . . .(1)

In second case,

3K = ${{2hc} \over \lambda }$ $-$ $\phi$0           . . .(2)

$\Rightarrow$   $3\left( {{{hc} \over \lambda } - {\phi _0}} \right)$ = ${{2hc} \over \lambda } - {\phi _0}$

$\Rightarrow$   ${{3hc} \over \lambda }$ $-$ 3$\phi$0 = ${{2hc} \over \lambda }$ $-$ $\phi$0

$\Rightarrow$   ${{hc} \over \lambda }$ = 2$\phi$0

$\Rightarrow$   $\phi$  =  ${{hc} \over {2\lambda }}$
3

### JEE Main 2017 (Offline)

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X–rays. It produces continuous as well as characteristic X-rays. If $\lambda$min is the smallest possible wavelength of X-ray in the spectrum, the variation of log$\lambda$min with log V is correctly represented in:
A
B
C
D

## Explanation

In X-ray tube, ${\lambda _{\min }} = {{hc} \over {eV}}$

$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$

Clearly, log ($\lambda$min) versus log V graph slope is negative hence option (3) is correct.
4

### JEE Main 2017 (Offline)

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths ${\lambda _A}$ to ${\lambda _B}$ after the collision is:
A
${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}$
B
${{{\lambda _A}} \over {{\lambda _B}}} = 2$
C
${{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}$
D
${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}$

## Explanation

From question, mA = M; mB = ${m \over 2}$

uA = V and uB = 0

Let after collision velocity of A = V1 and

velocity of B = V2

Applying law of conservation of momentum,

mu = mv1 + $\left( {{m \over 2}} \right){v_2}$

$\Rightarrow$ 24= 2v1 + v2 ........(i)

By law of collision

$e = {{{v_2} - {v_1}} \over {u - 0}}$

$\Rightarrow$ u = v2 - v1 ..........(ii)

[As collision is elastic, e = 1]

using eqns (i) and (ii)

v1 = ${4 \over 3}$ and v2 = ${4 \over 3}v$

We know, de-Broglie wavelength $\lambda$ = ${h \over p}$

$\therefore$ ${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$ = ${{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}$ = 2