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Updated on : 2022-09-05

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Correct option is D)

From $E=W_{0}+21 mv_{max}⇒v_{max}=m2E −m2W_{0} $

where $E=λhc $

If wavelength of incident light charges from $λ$ to $43λ $ (Decreases)

Let energy of incident light charges from $E$ and speed of fastest electron changes from $v$ to $v_{′}$ then

$v_{′}=m2E_{′} −m2W_{0} $

As $E∝λ1 ⇒E_{′}34 E$

Hence $v_{′}=m2(34 )E −m2W_{0} $

$⇒v_{′}=(4/3)_{1/2}m2E −m(4/3)_{1/2}2W_{0} $

So, $v_{′}>(4/3)_{1/2}v$

Video Explanation

Solve any question of Dual Nature of Radiation And Matter with:-

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