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Answer 1 · 2017-02-28T12:10:21

By photoelectric equation of Einstein we have

$E = ϕ_{o} + K_{max}$ $E=phi_o + K_"max"$

Where
$ϕ_{o} \to photoelectric work function$ $phi_o ->"photoelectric work function"$

$E \to enrgy of photon absorbed$ $E-> "enrgy of photon absorbed"$

$K_{max} \to maximum KE of emitted electron$ $K_"max"-> "maximum KE of emitted electron"$

Inserting $ϕ_{o} = 1.9 e v and E = 2.1 e v$ $phi_o = 1.9ev and E=2.1ev$ in the equation we get

$2.1 = 1.9 + K_{max}$ $2.1=1.9+ K_"max"$

$\Rightarrow K_{max} = 0.2 e v$ $=>K_"max"=0.2ev$

If KE is doubled then the value of E will be E'

$E'=phi_o + 2xxE_"max"$

$E'=1.9+2xx0.2=2.3ev$

So the energy of photon to be enhanced by $(E')/E=2.3/2.1=23/21$ times to get the KE of photoelectron doubled.