Q155 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Physics

The de-Broglie wavelength of an electron with kinetic energy of $320 \mathrm{eV}$ is $\left(\right.$ Take $\mathrm{h}=6.0 \times 10^{-34}$ SI unit, mass of electron $=\mathrm{m}_{\mathrm{e}}$ $=9.0 \times 10^{-31} \mathrm{~kg}$. Charge of an electron $=1.6 \times 10^{-19} \mathrm{C}$ )

  • A
    85.8 pm
  • B
    110.5 pm
  • Correct answer
    C
    62.5 pm
  • D
    50 pm

Solution

We have $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}} \\ & =\frac{6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 320 \times 1.6 \times 10^{-19}}} \\ & =6.25 \times 10^{-11} \mathrm{~m}=62.5 \mathrm{pm}\end{aligned}$

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

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