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A horizontal cesium plate(phi=1.9eV) is ...

A horizontal cesium plate`(phi=1.9eV)` is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250nm and above. What should be the minimum value of v so that the vertically upward component of velocity is nonpositive for each photoelectron?

Text Solution

Verified by Experts

Minimum energy of the photon falling on photoelectric plate
`=hv=(hc)/(lamda)=((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-19))`
According to Einstein's photoelectric equation,
`(1)/(2)mv^(2)=(hc)/(lamda)-W`
`[v_(min)]=(2)/(m)[(hc)/(lamda)-W]`
`=(2)/(9.1xx10^(-31))[((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-9))-1.9xx(1.6xx10^(19))]`
`=1.08xx10^(12)`
`v_(min)=1.04xx10^(6)ms^(-1)`
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