QJEEM18Ph48bf: JEE Main 2018 Physics - Class 12 JEEM18Ph48bf Solution: The de-Broglie wavelength ( B ) associated with the electron…

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 a₀ is radius of Bohr's orbit having value (0.53) 1² $\mathop A\limits^o$ = 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,

${\lambda _G} = {{2\pi {a_0} \times 1} \over 1} = 2\pi {a_0}$ and ${\lambda _B} = {{2\pi {a_0} \times 3} \over 1} = 6\pi {a_0}$

Then, ${\lambda _B} = 3{\lambda _G}$