The ratio of de-Broglie wavelengths of two pa | vedclass

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(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since $p = \sqrt{2mK}$,where $m$ is mass and $K$ is kinetic energy,the ratio of wave

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since $p = \sqrt{2mK}$,where $m$ is mass and $K$ is kinetic energy,the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = \sqrt{\frac{2m_2 K_2}{2m_1 K_1}}$.Given $\frac{m_1}{m_2} = \frac{1}{3}$ and $\frac{K_1}{K_2} = \frac{2}{1}$,we have $\frac{m_2}{m_1} = 3$ and $\frac{K_2}{K_1} = \frac{1}{2}$.Substituting these values: $\frac{\lambda_1}{\lambda_2} = \sqrt{3 	imes \frac{1}{2}} = \sqrt{\frac{3}{2}} = \sqrt{3}:\sqrt{2}$.

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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