First slide

Matter waves (DC Broglie waves)

Question

According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is $10^{- 9}$ . The principle quantum number for this electron is

Moderate

A

1
B

2
C

3
D

4

Solution

$m v r_{n} = n \frac{h}{2 π}, 2 {πr}_{n} = nλ, 2 π (0.53 X 10^{- 10}) \frac{n^{2}}{Z} = nλ (z = 1), n = \frac{λ}{2 π X 0.53 X 10^{- 10}} = \frac{10^{- 9}}{2 \times 3.14 \times 0.53 \times 10^{- 10}} = 3$

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