Question

For the Bohr's first orbit of circumference 2πr , the de- Broglie wavelength of revolving electron will be
(a) 2πr
(b) πr
(c) $\frac{1}{2\pi r}$
(d) $\frac{1}{4\pi r}$

Answer 1

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Answer 2

Answer: The correct answer is 2πr.

Explanation:

The formula for the De- Broglie wavelength of revolving electron is as follows;

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

Here, h is Planck's constant, m is the mass of the electron and v is the velocity of the electron.

Rearrange the above expression.

$mv=\frac{h}{\lambda }$                               ......... (1)

It is given in the problem that Bohr's first orbit of circumference is 2πr.

The expression from Bohr model is as follows;

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

Here, n is the integral multiple.

Combine (1) and (2).

$2\pi(r)=n\lambda$

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

$mvr=\frac{h}{2\pi}$

$2\pi(r)=\frac{h}{mv}$

Therefore, the correct option is (d).