1

### JEE Main 2017 (Online) 9th April Morning Slot

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are :

[Take Planck's constant h $=$ 6.62 $\times$ 10$-$34 Js]
A
1020
B
1018
C
1022
D
1019
2

### JEE Main 2018 (Online) 15th April Morning Slot

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are ${\lambda _1}$ and ${\lambda _2},$ their de Broglie wavelength in the frame of reference attached to their center of massis :
A
${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$
B
${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$
C
${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$
D
${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$

## Explanation

As we know,

momentum (p) = ${h \over \lambda }$

Let one perticle is moving in x direction and other is in y dirrection.

$\therefore\,\,\,\,$ momentum of each electrons ${h \over {{\lambda _1}}}\widehat i$ and ${h \over {{\lambda _2}}}\widehat j$

$\therefore\,\,\,\,$ Velocity of each electrons ${h \over {m{\lambda _1}}}\widehat i$ and ${h \over {m{\lambda _2}}}\widehat j$

$\therefore\,\,\,\,$ Velocity of center of mass $\left( {{{\overrightarrow \upsilon }_{_{CM}}}} \right)$ = ${{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}$

Now, velocity of first electron about center of mass,

${\overrightarrow \upsilon _{_{1CM}}}$ = $\upsilon _1^{\widehat i}$ $-$ $\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} \right)$

= ${{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}$

Similarly,

${\overrightarrow \upsilon _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}$

Here,

$\left| {{{\overrightarrow \upsilon }_{_{1CM}}}} \right| = \left| {{{\overrightarrow \upsilon }_{_{2CM}}}} \right| = {1 \over 2}$

$\therefore\,\,\,\,$ $\upsilon$ = ${1 \over 2}$ $\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}}$

$\Rightarrow $$\,\,\,\, m \upsilon = {1 \over 2} {\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} } \Rightarrow$$\,\,\,\,$ ${h \over {{\lambda _{CM}}}}$ = h${\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }$

$\Rightarrow $$\,\,\,\, {1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}} \Rightarrow$$\,\,\,\,$ ${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$
3

### JEE Main 2018 (Online) 15th April Evening Slot

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal, then the ratio of velocities of the proton and the $\alpha$-particle will be :
A
4 : 1
B
2 : 1
C
1 : 2
D
1 : 4

## Explanation

We know, ${\lambda _p} = {h \over {{p_p}}}$ = ${h \over {{m_p}{v_p}}}$

Similarly, ${\lambda _\alpha } = {h \over {{m_\alpha }{v_\alpha }}}$

Given, ${\lambda _p} = {\lambda _\alpha }$

$\Rightarrow$ ${h \over {{m_p}{v_p}}} = {h \over {{m_\alpha }{v_\alpha }}}$

$\therefore$ ${{{v_p}} \over {{v_\alpha }}} = {{{m_\alpha }} \over {{m_p}}}$ = ${{4{m_p}} \over {{m_p}}} = {4 \over 1}$
4

### JEE Main 2018 (Online) 16th April Morning Slot

The de-Broglie wavelength ($\lambda$B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($\lambda$G) by :
A
$\lambda$B = 2$\lambda$G
B
$\lambda$B = 3$\lambda$G
C
$\lambda$B = $\lambda$G/2
D
$\lambda$B = $\lambda$G/3