QJEEM21Chfd80: JEE Main 2021 Chemistry - Class 11 JEEM21Chfd80 Solution: A proton and a Li 3+ nucleus are accelerated by the same…

Given, mass of Li³⁺ = 8.3 times of mass of proton formula,

De-Broglie wavelength, $\lambda = {h \over {\sqrt {2mqV} }}$

Here, h = Planck's constant = 6.624 $\times$ 10 ^{$-$ 34} J-s

m = Mass of atom

q = Charge (or number of electrons)

${\lambda _{Li}} = {h \over {\sqrt {2{m_{Li}}(3e)v} }}$ .... (i)

${\lambda _P} = {h \over {\sqrt {2{m_P}(e)v} }}$ .... (ii)

Now, Eq. (i) divided by Eq. (ii), we get,

${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P}(e)v} \over {{m_{Li}}(3e)v}}}$

We know that m_Li = 8.3 m_p

${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$ = 2 $\times$ 10 ^{$-$ 1}

Hence, x $\times$ 10 ^{$-$ 1}, x = 2