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Q.

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27^oC and 127^oC respectively is

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a

$frac{1}{2}$

b

$sqrt {frac{3}{8}}$

c

$sqrt {frac{8}{3}}$

d

1

answer is C.

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Detailed Solution

de-Broglie wavelength $lambda = frac{h}{{m{v_{rms}}}}$ , rms velocity of a gas particle at the given temperature (T) is given as
$frac{1}{2}mv_{rms}^2 = frac{3}{2}kT Rightarrow {v_{rms}} = sqrt {frac{{3,kT}}{m}} Rightarrow m{v_{rms}} = sqrt {3,mk;T}$
$therefore lambda = frac{h}{{m{v_{rms}}}} = frac{h}{{sqrt {3,mkT} }}$
$Rightarrow frac{{{lambda _H}}}{{{lambda _{He}}}} = sqrt {frac{{{m_{He}}{T_{He}}}}{{{m_H}{T_H}}}} = sqrt {frac{{4,(273 + 127)}}{{2,(273 + 27)}}} = sqrt {frac{8}{3}}$

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The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27oC and 127oC respectively is