Question 51
The de-Broglie's wavelength of an electron in the 4th orbit is ______ $$\pi a_0$$. ($$a_0$$ = Bohr's radius)
Correct Answer: 8
Solution
For the nth orbit, the de Broglie condition states:
$$2\pi r_n = n\lambda$$
For n = 4: $$\lambda = \frac{2\pi r_4}{4}$$
The radius of the nth orbit: $$r_n = n^2 a_0$$, so $$r_4 = 16a_0$$
$$\lambda = \frac{2\pi \times 16a_0}{4} = 8\pi a_0$$
The answer is 8.
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