AP EAMCET · Chemistry · Structure of Atom
The de Broglie wavelength of an electron with kinetic energy of 2.5 eV is (in m )
\(\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}, m_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}\right)\)
- A \(\frac{h \times 10^{-25}}{\sqrt{72}}\)
- B \(\frac{h \times 10^{25}}{\sqrt{72}}\)
- C \(\frac{\sqrt{72}}{h \times 10^{-25}}\)
- D \(\frac{\sqrt{72}}{h \times 10^{25}}\)
Answer & Solution
Correct Answer
(B) \(\frac{h \times 10^{25}}{\sqrt{72}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{KE}=\frac{1}{2} \mathrm{~m} v^2\) \(v=\sqrt{\frac{2 \cdot \mathrm{KE}}{\mathrm{~m}}}\) de broglie wavelength \((\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}\)…
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