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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minPYQ_2015

PhysicsMediumsingle choice

Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The de Broglie wavelength of the emitted electron is:NEET 2015 (Phase 2)

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Answer:

Solution:

$K E_{m a x} = \frac{h c}{λ} - Ψ$
$\Rightarrow K E_{m a x} = \frac{1240}{500} - 2.28$
$\Rightarrow K E_{m a x} = 2.48 - 2.28 = 0.2 e V$
$λ_{m i n} = \frac{h}{\sqrt{2 m {(KE)}_{\max}}} = \frac{\frac{20}{3} \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ_{m i n} = \frac{25}{9} \times 10^{- 9} = 2.80 \times 10^{- 9} m$
So, $λ \geq 2.8 \times 10^{- 9} m$ .

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

⏱ 2mℹ️ Source: PYQ_2015

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