AP EAMCET · Chemistry · Structure of Atom
The velocity of the electron in Bohr's first orbit is \(x \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\). The de-Broglie wavelength associated with it (in nm) is \(\left(m_e=9 \times 10^{-31} \mathrm{~kg}, h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\)
- A \(\frac{x}{1.43}\)
- B \(\frac{x}{0.73}\)
- C \(\frac{0.73}{x}\)
- D \(\frac{0.073}{x}\)
Answer & Solution
Correct Answer
(C) \(\frac{0.73}{x}\)
Step-by-step Solution
Detailed explanation
According to de-Broglie's equation, \(\lambda=\frac{h}{m_e v}\) \(\begin{aligned} & \lambda=\text { de-Broglie wavelength } \\ & h=\text { Planck's constant }=6.626 \times 10^{-34} \mathrm{JS}\end{aligned}\)…
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