Atoms And Nuclei Question 120
Question: The de-Broglie wavelength of an electron in the first Bohr orbit is [KCET 2002]
Options:
A) Equal to one fourth the circumference of the first orbit
B) Equal to half the circumference of the first orbit
C) Equal to twice the circumference of the first orbit
D) Equal to the circumference of the first orbit
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Answer:
Correct Answer: D
Solution:
$ mvr_{n}=\frac{nh}{2\pi }\Rightarrow pr_{n}=\frac{nh}{2\pi }\Rightarrow \frac{h}{\lambda }\times r_{n}=\frac{nh}{2\pi } $
$ \Rightarrow \lambda =\frac{2\pi r_{n}}{n}, $ for first orbit $ n=1 $ so $ \lambda =2\pi r_{1} $ = circumference of first orbit