JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 15)
The de Broglie wavelength of an electron in the
4th Bohr orbit is :
2$$\pi $$a0
6$$\pi $$a0
8$$\pi $$a0
4$$\pi $$a0
Explanation
According to Bohr’s model
$${r_n} = {{{n^2}} \over Z} \times {a_0}$$
Also 2$$\pi $$$${r_n}$$ = n$$\lambda $$
$$ \Rightarrow $$ 2$$\pi $$$${{{n^2}} \over Z} \times {a_0}$$ = n$$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = $$2\pi \times {n \over Z} \times {a_0}$$
For n = 4 and Z = 1
$$ \Rightarrow $$ $$\lambda $$ = 8$$\pi $$$${a_0}$$
$${r_n} = {{{n^2}} \over Z} \times {a_0}$$
Also 2$$\pi $$$${r_n}$$ = n$$\lambda $$
$$ \Rightarrow $$ 2$$\pi $$$${{{n^2}} \over Z} \times {a_0}$$ = n$$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = $$2\pi \times {n \over Z} \times {a_0}$$
For n = 4 and Z = 1
$$ \Rightarrow $$ $$\lambda $$ = 8$$\pi $$$${a_0}$$
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