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PhysicsModern PhysicsPhotoelectric effectEasy2 minPYQ_2020

PhysicsEasysingle choice

A light of wavelength $4000 \overset{\circ}{A}$ is allowed to fall on a metal surface having work function $2 eV$ . The maximum velocity of the emitted electrons is $(h = 6.6 \times 10^{- 34} J s)$

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Answer:

Solution:

$\frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ (i n e V)$

$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 2$

$= 3.1 - 2 = 1.1 e V = 1.1 \times 1.6 \times 10^{- 19} J$

$= 1.76 \times 10^{- 19} J$

$v = \frac{1.76 \times 10^{- 19} \times 2}{9 \times 10^{- 13}}$

$= 6.2 \times 10^{5} m s^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2020

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