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

Answers (1)

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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