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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2015

PhysicsEasystatement

The de-Broglie wavelength associated with the electron in the $n = 4$ level is:

Options:

Answer:

Solution:

De-Broglie wavelength is given as

$λ = \frac{h}{p}$

$∵ \frac{p^{2}}{2 m} = k \Rightarrow p = \sqrt{2 m k}$

$\Rightarrow λ = \frac{h}{\sqrt{2 m k}}$

At ground level energy $= E_{0}$ so

At $n = 4$ level energy $= \frac{E_{0}}{n^{2}} = \frac{E_{0}}{16}$

$∴$ at ground level

$λ_{0} = \frac{h}{\sqrt{2 m E_{0}}}$

At n = 4 level

$λ^{'} = \frac{h}{\sqrt{2 m \frac{E_{0}}{16}}} = \frac{4 h}{\sqrt{2 m E_{0}}}$

$\Rightarrow λ^{'} = 4 λ_{0}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: PYQ_2015

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