CBSE Class 12 · 2018 · Modern PhysicsShort Answer

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If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? (Na: 1.92 eV; K: 2.15 eV; Ca: 3.20 eV; Mo: 4.17 eV.)

Answer: Photon energy $\approx 3.0 eV$ ; only Na and K ( $ϕ_{0} < 3 eV$ ) emit.

Solution

Energy of each incident photon:

E = \frac{h c}{λ} = \frac{1240 eV nm}{412.5 nm} \approx 3.0 eV .

Photoemission occurs only when

E \geq ϕ_{0}

. Here

E = 3.0 eV

exceeds the work functions of Na (

1.92 eV

) and K (

2.15 eV

) but is less than those of Ca (

3.20

) and Mo (

4.17

Na and K show photoelectric emission

CBSE Class 12 Board · 2018 · Subjective

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