Dual Nature of Matter

JEE Main level, single correct, medium

77 of 200

Question 77 of 200

The de-Broglie wavelength (λB)\left({\lambda }_{B}\right) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λG)\left({\lambda }_{G}\right) by:

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Answer: (A) λB=3λG{\lambda }_{B}=3{\lambda }_{G}

We know that wavelength is inversely proportional to momentum, let λB{\lambda }_{B} is de-Broglie Wavelength corresponds to momentum PB{P}_{B}, similarly let λG{\lambda }_{G} is Wavelength at ground state corresponds to momentumPG{P}_{G},
λBλG=PGPB=mvGmvB\dfrac{{\lambda }_{B}}{{\lambda }_{G}}=\dfrac{{P}_{G}}{{P}_{B}}=\dfrac{m{v}_{G}}{m{v}_{B}}, and velocity, V∝znV\propto \dfrac{z}{n}, where ZZ is the atomic number, and nn is a number of orbits, here, nG=1{n}_{G}=1 (ground state), nB=3{n}_{B}=3 (second excited state). So, λBλG=nBnG=31\dfrac{{\lambda }_{B}}{{\lambda }_{G}}=\dfrac{{n}_{B}}{{n}_{G}}=\dfrac{3}{1}⇒λB=3λG\Rightarrow {\lambda }_{B}=3{\lambda }_{G}

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