Question 210: Dual Nature of Matter and Radiation

Question

The de-Broglie wavelength ($$\lambda $$B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda $$G) by :

Accepted Answer

We know that, $$\lambda  = {h \over {mv}}$$
From third Bohr's postulate, we have
$$mvr = n{h \over {2\pi }}$$
$${h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda  = {{2\pi r} \over n}$$
Since, $$r = {a_0}{{{n^2}} \over Z}$$, where a
0
is radius of Bohr's orbit having value (0.53) 1
2
$$\mathop A\limits^o $$  = 0.53 $$\mathop A\limits^o $$, therefore,
$$\lambda  = {{2\pi {a_0}{n^2}} \over {n\,.\,Z}} = {{2\pi {a_0}} \over Z}\,.\,n$$
For Hydrogen Z = 1. Therefore,
$${\lambda _G} = {{2\pi {a_0} 	imes 1} \over 1} = 2\pi {a_0}$$ and $${\lambda _B} = {{2\pi {a_0} 	imes 3} \over 1} = 6\pi {a_0}$$
Then, $${\lambda _B} = 3{\lambda _G}$$

Answer

$$\lambda $$B = 2$$\lambda $$G

Answer

$$\lambda $$B = 3$$\lambda $$G

Answer

$$\lambda $$B = $$\lambda $$G/2

Answer

$$\lambda $$B = $$\lambda $$G/3

Dual Nature of Matter and Radiation

The de-Broglie wavelength ($\lambda B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (\lambda $G) by :

Choose the correct answer: