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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}$.For a gas molecule,the average kinetic energy $E_k = \frac{

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$1:1$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}$.For a gas molecule,the average kinetic energy $E_k = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.Thus,$\lambda = \frac{h}{\sqrt{2m(\frac{3}{2}kT)}} = \frac{h}{\sqrt{3mkT}}$.This implies $\lambda \propto \frac{1}{\sqrt{mT}}$.For oxygen $(O_2)$,molar mass $M_O = 32 \ g/mol$ and temperature $T_O = 21 + 273 = 294 \ K$.So,$m_O T_O = 32 	imes 294 = 9408$.For nitrogen $(N_2)$,molar mass $M_N = 28 \ g/mol$ and temperature $T_N = 63 + 273 = 336 \ K$.So,$m_N T_N = 28 	imes 336 = 9408$.Since the product $mT$ is the same for both,the ratio $\frac{\lambda_O}{\lambda_N} = \sqrt{\frac{m_N T_N}{m_O T_O}} = \sqrt{\frac{9408}{9408}} = 1$.Therefore,the ratio is $1:1$.

Find the ratio of the de Broglie wavelength of an oxygen molecule at $21^oC$ to that of a nitrogen molecule at $63^oC$.

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