The de Broglie wavelength of a particle X mov | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For particle $X$: $\lambda_X = \frac{h}{m_X v_X} = 1 \ \mathring{A}$.For particle $

Vedclass · Accepted Answer

$5.33$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For particle $X$: $\lambda_X = \frac{h}{m_X v_X} = 1 \ \mathring{A}$.For particle $Y$: $m_Y = 0.25 m_X$ and $v_Y = 0.75 v_X$.Therefore,$\lambda_Y = \frac{h}{m_Y v_Y} = \frac{h}{(0.25 m_X)(0.75 v_X)} = \frac{h}{0.1875 m_X v_X}$.Substituting $\lambda_X = \frac{h}{m_X v_X} = 1$,we get $\lambda_Y = \frac{1}{0.1875} = \frac{1}{3/16} = \frac{16}{3} \approx 5.33 \ \mathring{A}$.

The de Broglie wavelength of a particle $X$ moving with a certain velocity is $1 \ \mathring{A}$. If the mass of $Y$ is $25\%$ of $X$ and its velocity is $75\%$ of $X$,then the de Broglie wavelength of $Y$ will be ........... $\mathring{A}$.

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