Question

The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 eV) will be

• A. $4\times {10}^{10}Hz$
• B. $4\times {10}^{11}Hz$
• C. $4\times {10}^{14}Hz$
• D. $4\times {10}^{12}Hz$

Answer 1

The correct option is C $4\times {10}^{14}Hz$

Threshold wavelength ${\lambda }_0=\frac{12375}{W_0\left(eV\right)}=\frac{12375}{1.65}=7500\stackrel{\circ }{A}$
$\therefore$ so minimum frequency
$v_0=\frac{c}{{\lambda }_0}=\frac{3\times {10}^8}{7500\times {10}^{-10}}=4\times {10}^{14}Hz$.