MCQ
When the frequency of light incident on a metallic plate is double, the $KE$ of the emitted photoelectrons will be :
- ADoubled
- BHalved
- ✓Increased but more than doubled of the previous $KE$
- DUnchanged
$2 hv _1= hv _0+ KE _2 \ldots(2)$
By dividing equations $(1)$ and $(2)$, we get,
$KE _2=2 KE _1+ hv _0$
The value of kinetic energy will increase more than double of the previous $K.E.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.