Question 190: 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: