In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals and are , and , respectively, and they are related by In this experiment, the same source of monochromatic light is used for metals and while a different sourc

Question

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_{P}$ , $E_{Q}$ and $E_{R}$ , respectively, and they are related by $E_{P} = 2 E_{Q} = 2 E_{R} .$ In this experiment, the same source of monochromatic light is used for metals $P$ and $Q$ while a different source of monochromatic light is used for the metal $R$ . The work functions for metals $P, Q$ and $R$ are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal $R$ , in eV, is ________. [2021]

Smart Achievers · Accepted Answer

(6)

From photoelectric equation, $E = ϕ_{0} + {K . E .}_{\max}$

For metals $P$ and $Q$ ,

$E_{1} = ϕ_{P} + {KE}_{P} \Rightarrow E_{1} - 4 = E_{P}$ and $R$

$E_{P} = 2 E_{Q}$ $(given)$

$∴$ $E_{1} - 4 = 2 (E_{1} - 4.5) \Rightarrow E_{1} = 5 eV$

$E_{1} - 4 = E_{P} \Rightarrow 5 - 4 = E_{P}$

$∴$ $E_{P} = 1 eV$ and $E_{Q} = E_{R} = 0.5 eV$

For metal $R$ , $E_{2} - ϕ_{R} = {KE}_{R}$

$∴$ $E_{2} - 5.5 = 0.5$

$∴$ $E_{2} = 0.5 + 5.5 = 6 eV$

Solution

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Solution

Ans. (6) From photoelectric equation, E=ϕ0+K.E.max For metals P and Q, E1=ϕP+KEP⇒E1-4=EP and E1-4.5=EQ EP=2EQ (given) ∴ E1-4=2(E1-4.5)⇒E1=5 eV E1-4=EP⇒5-4=EP ∴ EP=1 eV and EQ=ER=0.5 eV For metal R, E2-ϕR=KER ∴ E2-5.5=0.5 ∴ E2=0.5+5.5=6 eV

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