The de Broglie wavelength of an electron in the 4th Bohr orbit is :
Solution
According to Bohr’s model
<br><br>${r_n} = {{{n^2}} \over Z} \times {a_0}$
<br><br>Also 2$\pi$${r_n}$ = n$\lambda$
<br><br>$\Rightarrow$ 2$\pi$${{{n^2}} \over Z} \times {a_0}$ = n$\lambda$
<br><br>$\Rightarrow$ $\lambda$ = $2\pi \times {n \over Z} \times {a_0}$
<br><br>For n = 4 and Z = 1
<br><br>$\Rightarrow$ $\lambda$ = 8$\pi$${a_0}$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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