De-Broglie wavelength of an electron orbiting in the \(n=2\) state of ...

Hint: \(2\pi r_n = n\lambda\) Step: Find the de-Broglie wavelength of the electron orbiting in the state \(n=2\) . According to de-Broglie hypothesis, \(2\pi r_n = n\lambda~~~...(1)\) According to Bohr's model for the hydrogen atom, the radius of the nucleus is given by; \(r_n = n^2 a_0~~~...(2)\) From equation \((1)\) and \((2)\) we get; \(\Rightarrow \lambda = 2\pi n a_0\) For \(n = 2\) , \(\Rightarrow \lambda = 4\pi a_0\) \(\Rightarrow \lambda = 4\times 3.14 \times 0.052= 0.65312~\text{nm}\) \(\Rightarrow \lambda \approx 0.67~\text{nm}\) Hence, option (4) is the correct answer.

De Broglies Explanation Of Bohrs Second Postulate

Hint:

2%5Cpi%20r_n%20%3D%20n%5Clambda

Step: Find the de-Broglie wavelength of the electron orbiting in the state

n%3D2

.
According to de-Broglie hypothesis,

2%5Cpi%20r_n%20%3D%20n%5Clambda~~~...(1)

According to Bohr's model for the hydrogen atom, the radius of the nucleus is given by;

r_n%20%3D%20n%5E2%20a_0~~~...(2)

From equation

(1)

and

(2)

we get;

%5CRightarrow%20%5Clambda%20%3D%202%5Cpi%20n%20a_0

For

n%20%3D%202

%5CRightarrow%20%5Clambda%20%3D%204%5Cpi%20a_0

%5CRightarrow%20%5Clambda%20%3D%204%5Ctimes%203.14%20%5Ctimes%200.052%3D%200.65312~%5Ctext%7Bnm%7D

%5CRightarrow%20%5Clambda%20%5Capprox%200.67~%5Ctext%7Bnm%7D

Hence, option (4) is the correct answer.

1.	$1.67~%20%5Ctext%7Bnm%7D$	2.	$2.67~%20%5Ctext%7Bnm%7D$
3.	$0.067~%20%5Ctext%7Bnm%7D$	4.	$0.67~%20%5Ctext%7Bnm%7D$

De Broglies Explanation Of Bohrs Second Postulate

High-Yield Trend

Questions
1 MCQs

Official Solution

De Broglies Explanation Of Bohrs Second Postulate

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs