Q128 / 160 · TS EAMCET 2017 · Physics

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \mathrm{~nm}$ to $0.5 \mathrm{~nm}$ is

  • A
    four-times the initial energy
  • B
    equal to the initial energy
  • C
    two-times the initial energy
  • Correct answer
    D
    three-times the initial energy

Solution

We know that, We have, $\lambda=\frac{h}{\sqrt{2 m k}}$ $\therefore$ Kinetic energy $(K E) \propto \frac{1}{\lambda^2}$ Hence, $$ \frac{K E_2}{K E_1}=\left(\frac{\lambda_1}{\lambda_2}\right)^2 $$ Here, $\lambda_1=1 \mathrm{~nm}$ $$ \begin{aligned} \lambda_2 & =0.5 \mathrm{~nm} \\ \frac{K E_2}{K E_1} & =\left(\frac{1}{0.5}\right)^2 \\ K E_2 & =4 K E_1 \end{aligned} $$ Hence, the kinetic energy is increases three times. $$ \triangle K E=3 K E_1 $$

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

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