TS EAMCET · Physics · Dual Nature of Matter
The potential energy of an electron in an orbit of hydrogen atom is -6.8 eV .
The de Broglie wavelength of the electron in this orbit is
( \(r_0\) is Bohr radius)
- A \(2 \pi r_o\)
- B \(4 \pi r_o\)
- C \(\pi r_o\)
- D \(3 \pi r_o\)
Answer & Solution
Correct Answer
(B) \(4 \pi r_o\)
Step-by-step Solution
Detailed explanation
\(U_n = 2 E_n = 2 \left( \frac{-13.6}{n^2} \right)\) \(-6.8 = \frac{-27.2}{n^2} \implies n^2 = \frac{-27.2}{-6.8} = 4 \implies n = 2\) \(2 \pi r_n = n \lambda\) \(2 \pi (n^2 r_o) = n \lambda\) \(\lambda = 2 \pi n r_o\) \(\lambda = 2 \pi (2) r_o = 4 \pi r_o\)
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