In a photoelectric experiment, the graph of frequency v of incident light (in Hz) and stopping potential V (in V) is as shown in the figure. Planck's constant is (e is the elementary charge)

Options

A. `e(ab)/(bc)`
B. `e(cb)/(ac)`
C. `e(ac)/(bc)`
D. `e(ac)/(ab)`

Answer

A

Explanation

(a) : According to Einstein's photoelectric equation
`hu=(K.E)_(max)+phi_(0)`
where `phi_(0)` is the Planck's constant
Also, `(K.E.)_(max)=eV`
where e is the elementary charge, V is the stopping potential
`eV=hu-phi_(0)rArrV=(h)/(e)v-(phi_(0))/(e)`
Hence, the graph between V and v is a straight line and slope of this graph is given by
Slope `=(h)/(e)` . . . (i)
From the graph in the question Slope `=(ab)/(bc)` . . . (ii)
From (i) and (ii), we get, `h=e(ab)/(bc)`