The work function of a metallic surface is `5.01 eV`. The photo - electrons are emitted when light of wavelength `2000 Å` falls on it . The potential difference applied to stop the fastest photo - electrons is `[ h = 4.14 xx 10^(-15) eV sec]`

Options

A. 1.2 volts
B. 2.24 volts
C. 3.6 volts
D. 4.8 volts

Answer

A

Explanation

`K_("max") =hv-phi_0`
`implies eV_0 = (hc)/lamda -phi_0`
`implies V_0 = (4.14 xx10^(-15) xx3 xx10^8)/(2000 xx10^(-10))-5.01`
`= 6.21 - 5.01 = 1.2V`