Single Choice

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

A$$4\times 0.529\overset{o}{A}$$
B$$2\pi \times 0.529\overset{o}{A}$$
Correct Answer
C$$\displaystyle\frac{0.529}{2\pi}\overset{o}{A}$$
D$$0.529\overset{o}{A}$$

Solution

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is $$ 2\pi \times 0.529\overset{o}{A}$$.

First Bohr orbit of 'H' atom has radius $$r=0.529 \: A^o$$

Also, the angular momentum is quantised.

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

$$2 \pi r=\dfrac {h}{mv} = \lambda$$

$$\lambda = 2\pi \times 0.529\overset{o}{A}$$

So, the correct option is $$B$$

SIMILAR QUESTIONS

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