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| 1. |
Light of frequency `7.21xx10^(14)Hz` is incident on a metal surface. Electrons with a maximum speed of `6.0xx10^(5)ms^(-1)` are ejected from the surface. What is the threshold frequency for photoemission of electrons? `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`. |
| Answer» `1/2mv_(max)^(2)=hv-hv_(0) or v_(0) =v-(mv_(max)^(2))/(2h)=7.21xx10^(14)-((9.1xx10^(-31))xx(6xx10^(5))^(2))/(2xx(6.63xx10^(-34)))=4.74xx10^(14) Hz` | |