The de Broglie wavelength and kinetic energy | vedclass

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(B) The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum and $h$ is Planck's constant.The kinetic ene

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$2$

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(B) The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum and $h$ is Planck's constant.The kinetic energy $K$ of a particle is related to its momentum $p$ by $K = \frac{p^2}{2m}$,where $m$ is the mass of the particle.From this,we can express the momentum as $p = \sqrt{2mK}$. Substituting this into the de Broglie wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.Since the mass $m$ of the particle remains constant,the relationship between wavelength and kinetic energy is $\lambda \propto \frac{1}{\sqrt{K}}$.Given $\lambda = 2000 \ \mathring{A}$ for $K = 1 \ 	ext{eV}$,and we need to find $\lambda'$ for $K' = 1 \ 	ext{MeV} = 10^6 \ 	ext{eV}$.Using the ratio: $\frac{\lambda'}{\lambda} = \sqrt{\frac{K}{K'}} = \sqrt{\frac{1 \ 	ext{eV}}{10^6 \ 	ext{eV}}} = \sqrt{\frac{1}{10^6}} = \frac{1}{10^3}$.Therefore,$\lambda' = \frac{\lambda}{10^3} = \frac{2000}{1000} \ \mathring{A} = 2 \ \mathring{A}$.

The de Broglie wavelength and kinetic energy of a particle are $2000 \ \mathring{A}$ and $1 \ \text{eV}$ respectively. If its kinetic energy becomes $1 \ \text{MeV}$,then its de Broglie wavelength becomes $...... \ \mathring{A}$.

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