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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2020

PhysicsEasysingle choice

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 nm$ to $0.5 nm$ is

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Answer:

Solution:

De-Broglie Wavelength

$λ = \frac{h}{\sqrt{2 mE}}$

$∴ \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$

$\Rightarrow \frac{1 \times 1 0^{- 9}}{0.5 \times 1 0^{- 9}} = \sqrt{\frac{E_{2}}{E_{1}}}$

$\Rightarrow 2 = \sqrt{\frac{E_{2}}{E_{1}}}$

$\Rightarrow \frac{E_{2}}{E_{1}} = 4$

$∴ E_{2} = 4 E_{1}$

$∴$ Energy to be added $= E_{2} - E_{1}$

$= 4 E_{1} - E_{1} = 3 E_{1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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