The de - Broglie wavelength associated with the electron in the n=4 level is : Option 1)  two times the de-Broglie wavelength of the electron in the ground state   Option 2)   four times the de-Broglie wavelength of the electron in the ground state Option 3) half of the de-Broglie wavelength of the electron in the ground state   Option 4) 1/4th of the de-Broglie wavelength of the electron in the ground state

As we learnt in

De - Broglie wavelength -

$\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}$

- wherein

$h= plank's\: constant$

$m= mass \: of\: particle$

$v= speed \: of\: the \: particle$

$E= Kinetic \: energy \: of \: particle$

From debroglie equation: $p=\frac{h}{\lambda}\ \; and\ \;mvr=\frac{nh}{2\pi}$

$\Rightarrow\ \;{2\pi r}=n{\lambda}$                        (1)

$r=r_{0}\frac{n^{2}}{z}$

$\Rightarrow\ \;2{\pi r_{0}}\frac{n^{2}}{z}=n{\lambda}$

$\Rightarrow\ \;{\lambda}=n.\frac{2{\pi r_{0}}}{z}$

In ground state n = r $\therefore\ \;{\lambda}=\frac{2{\pi r_{0}}}{z}$ in  n = 4,  ${\lambda}=\frac{8{\pi r_{0}}}{z}$

$\therefore$    Correct option is (2)

Option 1)

two times the de-Broglie wavelength of the electron in the ground state

This is an incorrect option.

Option 2)

four times the de-Broglie wavelength of the electron in the ground state

This is the correction.

Option 3)

half of the de-Broglie wavelength of the electron in the ground state

This is an incorrect option.

Option 4)

1/4th of the de-Broglie wavelength of the electron in the ground state

This is an incorrect option.

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