The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The sum of the slopes of the stopping potential versus frequency plot for Gold and Aluminimum is  $x\frac{h}{e},\text{\hspace{0.17em}then\hspace{0.17em}}x=\_\_\_\_\_.$

For photoelectric effect, eVs = KE_max ….. (i)
Where V_s is the stopping potential.
For incident wave of frequency  $\upsilon ,h\upsilon =\varphi +\mathrm{KE}\mathrm{.......}\left(ii\right)$
Where $\varphi$  is the work function. By equation (i) and (ii),  $e{V}_s=h\upsilon -\varphi$
$V_s=\left(\frac{h}{e}\right)V+\left(\frac{-\varphi }{e}\right)$  in this, slope $=\left(\frac{h}{e}\right)$  
As slope is independent of the value of work function, 

Slope of stopping potential of gold = Slope of stopping potential of aluminium. 

Sum  $=\frac{h}{e}+\frac{h}{e}=\frac{2h}{e}$