1.

The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength `2000A` falls on it. The potential difference applied to stop the fastes photoelectrons is `[h=4.14xx10^(-15)eVs]`A. 1.2 VB. 2.24 VC. 3.6 VD. 4.8 V

Answer» Correct Answer - A
`eV=hv-phi_0`
`=((12375)/(2000)-5.01)eV`
`V=(6.1875-5.01)V-1.18Vapprox1.2V`


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