Question

A photoelectric surface has a work function of 4 ev. What is the maximum velocity of the photoelectrons emitted by light of frequency 3 × 10¹⁵ Hz?

Answer 1

Photoelectric equation

E_k = h\(\nu\) - \(\phi_0\)

\(\frac12mv^2=h\nu-\phi_0\)

v² = \(\frac{2(h\nu-\phi_0)}m\) 

v = \(\sqrt{\frac{2(h\nu-\phi_0)}m}\) 

v = \(\sqrt{\frac{2(6.64\times10^{-34}\times3\times10^{15}-4ev)}{9.1\times10^{-31}}}\) 

v = \(\sqrt{\frac{2[(19.92\times10^{-19})-4ev]}{9.1\times10^{31}}}\) 

v = \(\sqrt{2.97\times10^6}\)

v = 1.72 x 10⁶ m/s