JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 16)
A proton and a Li3+ nucleus are accelerated by the same potential. If $$\lambda _{Li}$$ and $$\lambda _p$$ denote the
de Broglie wavelengths of Li3+ and proton respectively, then the value of
$${{{\lambda _{Li}}} \over {{\lambda _p}}}$$ is x $$ \times $$ 10-1.
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li3+ = 8.3 mass of proton]
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li3+ = 8.3 mass of proton]
Answer
2
Explanation
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 mLi = 8.3 mp
$${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$$ = 2 $$\times$$ 10$$-$$1
Hence, x $$\times$$ 10$$-$$1, x = 2
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 mLi = 8.3 mp
$${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$$ = 2 $$\times$$ 10$$-$$1
Hence, x $$\times$$ 10$$-$$1, x = 2
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