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Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

PhysicsEasysingle choice

If an electron drops from $4^{t h}$ orbit to $2^{n d}$ orbit in an $H - atom$ , then

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Answer:

Solution:

$E_{n} = - 13.6 \frac{z^{2}}{n^{2}}$

$∆ E = E_{4} - E_{2}$

$= - 13.6 z^{2} [\frac{1}{16} - \frac{1}{4}]$

$= - 13.6 \times \frac{1 - 4}{16} = \frac{13.6 \times 3}{16}$

$= 2.55 eV$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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