Photoelectrons ar ejected from a surface when light of wavelenght `lamda_1=550 nm` is incident on it. The stopping potential for such electrons is `lamda_(S1)=0.19V`. Suppose that radiation of wavelength `lamda_2=190nm` is incident of the surface.
Q. Calculate the stopping potential `V_(S2)`.

Options

A. `500xx10^(12)` Hz
B. `480xx10^(13)` Hz
C. `520xx10^(11)` Hz
D. `460xx10^(13)` Hz

Answer

A

Explanation

Fron Einstein's relation
`eV_S=hv-W`
As work function is a constant for a surface,
`e(V_(S2)-V_(S1)=h(v_2-v_1)`
`V_(S2)=V_(S1)+(h)/(e)(v_2-v_1)`
`=V_(S1)+(hc)/(e)((1)/(lamda_2)-(1)/(lamda_1))`
`=0.19+1240((1)/(190)-(1)/(15))=4.47 V`
`W=(hc)/(lamda_1)-eV_(S1)=(1240)/(550)-0.19=2.07 eV`
`hv_c=W`
`v_e=(W)/(h)=((2.07)(1.602xx10^(-19)))/(6.626xx10^(-34))approx500xx10^(12)Hz`