Q6 / 160 · AP EAMCET 2009 · Chemistry

The wavelengths of electron waves in two orbits is $3: 5$. The ratio of kinetic energy of electrons will be

  • Correct answer
    A
    $25: 9$
  • B
    $5: 3$
  • C
    $9: 25$
  • D
    $3: 5$

Solution

From de-Broglie's equation $\lambda=\frac{h}{m v}$ $\begin{aligned} \Rightarrow & \lambda^2 & =\frac{h^2}{m^2 v^2} \\ \Rightarrow & m v^2 & =\frac{h^2}{m \lambda^2}\end{aligned}$ $\begin{aligned} & \therefore & \mathrm{KE}(K) & =\frac{1}{2} m v^2 \\ & \therefore & \mathrm{KE}(K) & =\frac{1}{2} \frac{h^2}{m \lambda^2} \\ \Rightarrow & & \frac{K_1}{K_2} & =\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{5}{3}\right)^2 \\ & \therefore & K_1 & : K_2=25: 9\end{aligned}$

Hence, the correct answer is option: 'A'.

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