The de Broglie wavelengths of two fast moving | vedclass

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Question

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is m

Vedclass · Accepted Answer

$1: 1$

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is mass,and $K$ is kinetic energy. Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mK}$,which implies $K = \frac{h^2}{2m\lambda^2}$. Given $\lambda_X = 1 \ nm$,$\lambda_Y = 3 \ nm$,and $m_X = 9m_Y$. The ratio of kinetic energies is $\frac{K_X}{K_Y} = \frac{h^2}{2m_X\lambda_X^2} 	imes \frac{2m_Y\lambda_Y^2}{h^2} = \frac{m_Y}{m_X} 	imes \left(\frac{\lambda_Y}{\lambda_X}ight)^2$. Substituting the values: $\frac{K_X}{K_Y} = \frac{m_Y}{9m_Y} 	imes \left(\frac{3}{1}ight)^2 = \frac{1}{9} 	imes 9 = 1$. Thus,the ratio is $1: 1$.

The de Broglie wavelengths of two fast moving particles $X$ and $Y$ are $1 \ nm$ and $3 \ nm$ respectively. The mass of $X$ is nine times the mass of $Y$. The ratio of kinetic energies of $X$ and $Y$ is:

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