Question

The de-broglie's wavelength of electron present in first bohr orbit of ‘h' atom is

Answer 1

Answer:

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Explanation:

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Answer 2

The De-broglie's wavelength of electron present in first Bohr orbit of ‘h' atom is $2\pi\times0.529A^0$.

Given:

• Electron present in first Bohr orbit of 'H' atoms.

To find:

• De-broglie's wavelength of electron = ?

Calculation:

We know that in the Bohr model, the angular momentum is given as follows:

$mvr=\frac{nh}{2\times\pi}$ ................. (1)

where:

m- mass

v - Velocity

r - Radius

n - Orbital state

h - Plank's constant

Momentum 'P' is product of mass and velocity and its formula is

$P = \frac{h}{\lambda}$

Substituting the value of 'P' in equation (1) gives:

$mvr=\frac{nh}{2\times\pi}$

$Pr=\frac{nh}{2\times\pi}$        (∵ mv = p)

$\frac{h}{\lambda}\times r$ = $\frac{nh}{2\times\pi}$     (∵ $P = \frac{h}{\lambda}$)

As 'h' is common in both side it gets cancelled and the equation becomes

$\frac{r}{\lambda} = \frac{n}{2\pi}$

On cross multiplication it gives

$\lambda\times n=r\times2\pi$

As we required to find wavelength, the formula must be altered as shown:

∴ $\lambda = \frac{r\times2\pi}{n}$

Now substituting the required values gives the de-broglie's wavelength:

The standard radius (r) value for hydrogen atom is $0.529A^0$, and value of n is 1.

$\lambda = \frac{r\times2\pi}{n}$

$\lambda = \frac{0.529\times2\pi}{1}$

$\lambda = 0.529\times2\pi$

$\lambda = 2\pi\times0.529$

Conclusion:

Thus the de-broglie's wavelength of electron present in first Bohr orbit of ‘h' atom is calculated as $2\pi\times0.529A^0$.

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