Discussion Forum

The work functions   of silver and sodium are 4.6 and 2.3 eV, respectively The  ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is 

Answer:

Explanation:

$eV_{0}=hf-W$

$\therefore$               $V_{0}=\left(\frac{h}{e}\right)f-\frac{W}{e}$

V₀   versus f graph  is a  straight line with slope=   $\frac{h}{e}$= a universal constant.

Therefore the ratio of two slopes sholud be 1