Question

The de - broglie wavelength associated with the electron in the n=4 level is

Answer 1

The debroglie wavelength is given by the formula I`ve mentioned below but since the velocity isnt given we`ll have to find it using formula given by Bohr v=2•2x10^6(Z|n)=2•2x10^6h=6•6x10^-34m=9•1x10^-31(KGS)Substitute all the values in (lambda)=h|mv to get the answerDone

Answer 2

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Solve it, The de - Broglie wavelength associated with the electron in the n=4 level is :

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Solve it, The de - Broglie wavelength associated with the electron in the n=4 level is :

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

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Answers (2) 437 Views

P perimeter

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 option (2)

Option 1)

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

This is an incorrect option.