The de Broglie wavelength of an electron with kinetic energy of 2.5 eV is (in m ) $\left(1 \mathrm{eV}=1.6…

The de Broglie wavelength of an electron with kinetic energy of 2.5 eV is (in m ) $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}, m_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}\right)$
  1. $\frac{h \times 10^{-25}}{\sqrt{72}}$
  2. $\frac{h \times 10^{25}}{\sqrt{72}}$
  3. $\frac{\sqrt{72}}{h \times 10^{-25}}$
  4. $\frac{\sqrt{72}}{h \times 10^{25}}$

Solution

$\mathrm{KE}=\frac{1}{2} \mathrm{~m} v^2$ $v=\sqrt{\frac{2 \cdot \mathrm{KE}}{\mathrm{~m}}}$ de broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ $\begin{aligned} & \therefore \lambda=\frac{\mathrm{h}}{\sqrt{\mathrm{~m} \times 2 . \mathrm{KE}}} \\ & \lambda=\frac{\mathrm{h}}{\sqrt{9 \times 10^{-31} \times 2 \times 2.5 \times 1.6 \times 10^{-19}}} \\ & =\frac{\mathrm{h}}{\sqrt{72} \times \sqrt{10^{-50}}} \\ & \lambda=\frac{\mathrm{h}}{\sqrt{72}} \times 10^{25} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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