Q156 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics

If a proton is accelerated through a potential difference of $1000 \mathrm{~V}$, then its de-Broglie wavelength is (given, $m_p=1.67 \times 10^{-27} \mathrm{~kg}$, $\left.h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$

  • Correct answer
    A
    $9.1 \times 10^{13} \mathrm{~m}$
  • B
    $9.1 \times 10^{+13} \mathrm{~m}$
  • C
    $1.09 \times 10^{15} \mathrm{~m}$
  • D
    $1.09 \times 10^{+15} \mathrm{~m}$

Solution

$\lambda_{\text {proton }}=\frac{h}{p}=\frac{0.286}{\sqrt{V}}(Å)$ (This formula is for protons.) $$ \therefore \quad \lambda=\frac{0.286}{\sqrt{1000}} \times 10^{-10} \mathrm{~m} \approx 9.1 \times 10^{-13} \mathrm{~m} $$

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

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