The two particles A and B have de Broglie wav | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m KE}}$.Thus,the ratio of wavelengths is $\frac{\lambda_{A}}{\lambda_{B}} = \sqr

Vedclass · Accepted Answer

$25 : 4$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m KE}}$.Thus,the ratio of wavelengths is $\frac{\lambda_{A}}{\lambda_{B}} = \sqrt{\frac{m_{B} KE_{B}}{m_{A} KE_{A}}}$.Given $\lambda_{A} = 1 \ nm$,$\lambda_{B} = 5 \ nm$,and $m_{A} = 4 m_{B}$.Substituting these values: $\frac{1}{5} = \sqrt{\frac{m_{B}}{4 m_{B}} 	imes \frac{KE_{B}}{KE_{A}}}$.Squaring both sides: $\frac{1}{25} = \frac{1}{4} 	imes \frac{KE_{B}}{KE_{A}}$.Therefore,$\frac{KE_{A}}{KE_{B}} = \frac{25}{4}$.

The two particles $A$ and $B$ have de Broglie wavelengths $1 \ nm$ and $5 \ nm$ respectively. If the mass of $A$ is four times the mass of $B$,the ratio of kinetic energies of $A$ and $B$ would be

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