The de Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and 1 eV respectively, If i

Question

The de Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and 1 eV respectively, If its kinetic energy becomes 1 MeV , then its de Broglie wavelength is\

Professor PC Thomas Classes · Answer

de Broglie wavelength of a particle associated with its kinetic energy is\ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$\ $	herefore \frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sqrt{\mathrm{K}_{2}}}{\sqrt{\mathrm{~K}_{1}}}$\ or $\frac{2000 \mathrm{~A}^{\circ}}{\lambda_{2}}=\sqrt{\frac{1 	imes 10^{6} \mathrm{eV}}{1 	imes \mathrm{eV}}}=10^{3}$\ $\Rightarrow \lambda_{2}=\frac{2000}{10^{3}}=2 \mathrm{~A}^{\circ}$\