The ratio of de Broglie wavelengths of two pa | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m K E}}$.Given the mass ratio $\frac{m_1}{m_2} = \frac{1}{3}$ and kinetic energy

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$\sqrt{3} : \sqrt{2}$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m K E}}$.Given the mass ratio $\frac{m_1}{m_2} = \frac{1}{3}$ and kinetic energy ratio $\frac{K E_1}{K E_2} = \frac{2}{1}$.The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 	imes K E_2}{m_1 	imes K E_1}}$.Substituting the values: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{3 	imes 1}{1 	imes 2}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

The ratio of de Broglie wavelengths of two particles,having mass ratio $1 : 3$ and kinetic energy ratio $2 : 1$ is

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