Home
Class 12
PHYSICS
Which is the correct relation between de...

Which is the correct relation between de- Brogile wavelength of an electron in the `n^(th)` Bohr orbit and radius of the orbit R?

A

`lambda = n2piR`

B

`lambda = (2piR)/n`

C

`lambda = (4 pi R)/n`

D

`lambda = (2 pi R)/nh`

Text Solution

AI Generated Solution

The correct Answer is:
To find the correct relation between the de Broglie wavelength of an electron in the \(n^{th}\) Bohr orbit and the radius of the orbit \(R\), we can follow these steps: ### Step 1: Understand de Broglie Wavelength The de Broglie wavelength \(\lambda\) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle. ### Step 2: Express Momentum For an electron, the momentum \(p\) can be expressed as: \[ p = mv \] where \(m\) is the mass of the electron and \(v\) is its velocity. ### Step 3: Substitute Momentum into the Wavelength Formula Substituting the expression for momentum into the de Broglie wavelength formula, we get: \[ \lambda = \frac{h}{mv} \] ### Step 4: Use Bohr's Third Postulate According to Bohr's third postulate, the angular momentum \(L\) of the electron in the \(n^{th}\) orbit is quantized and given by: \[ L = mvr = n \frac{h}{2\pi} \] where \(r\) is the radius of the orbit and \(n\) is a positive integer (the principal quantum number). ### Step 5: Rearrange for Velocity From the angular momentum equation, we can express the velocity \(v\) as: \[ v = \frac{n h}{2\pi m r} \] ### Step 6: Substitute Velocity Back into the Wavelength Formula Now, substituting this expression for \(v\) back into the de Broglie wavelength formula: \[ \lambda = \frac{h}{m \left(\frac{n h}{2\pi m r}\right)} = \frac{h \cdot 2\pi m r}{n h} = \frac{2\pi r}{n} \] ### Step 7: Final Relation Thus, we arrive at the relation between the de Broglie wavelength \(\lambda\) and the radius \(r\) of the orbit: \[ \lambda = \frac{2\pi r}{n} \] ### Conclusion The correct relation between the de Broglie wavelength of an electron in the \(n^{th}\) Bohr orbit and the radius of the orbit \(R\) is: \[ \lambda = \frac{2\pi R}{n} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

The de-Broglie wavelength of an electron in the first Bohr orbit is

The de Broglie wavelength of an electron in the nth Bohr orbit is related to the radius R of the orbit as:

The de Broglie wavelength of an electron in the 3rd Bohr orbit is

The de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius ris given by

The de-Broglie wavelength of an electron in 4th orbit is (where, r=radius of 1st orbit)

The de-Broglie wavelength of an electron in 4th orbit is (where, r=radius of 1st orbit)

de-Broglie wavelength of an electron in the nth Bohr orbit is lambda_(n) and the angular momentum is J_(n) then

The de-Broglie wavelength lambda_(n) of the electron in the n^(th) orbit of hydrogen atom is

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

Determine wavelength of electron in 4th Bohr's orbit of hydrogen atom