Two fast moving particles $X$ and $Y$ are associated with de Broglie wavelengths $1 \mathrm{~nm}$ and $4…
- $3: 1$
- $9: 1$
- $5: 12$
- $16: 9$
Solution
$\frac{\lambda_{1}}{\lambda_{2}}=\frac{m_{2} v_{2}}{m_{1} v_{1}} ; \frac{1}{4}=\frac{1}{9} \times \frac{v_{2}}{v_{1}}$
$\frac{v_{2}}{v_{1}}=\frac{9}{4} ; \frac{v_{1}}{v_{2}}=\frac{4}{9}$
$\mathrm{KE}=\frac{1}{2} m v^{2}$
$\frac{K E_{1}}{K E_{2}}=\frac{m_{1}}{m_{2}} \times \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{9}{1} \times\left(\frac{4}{9}ight)^{2}=\frac{16}{9}$
Asked in: JEE-TOPICTESTS-CHEMISTRY