Figure shows the stopping potential versus frequency $ϑ$ for photo electric emission from two metals A and B . Choose the correct statement (s) from the following :

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work function of A is greater than of B

the slope $\tan θ = \frac{h ϑ}{e}$

the ratio of intercepts on negative Y-axis is $\frac{C_{1}}{C_{2}} = \frac{ϕ_{A}}{ϕ_{B}}$ where $ϕ_{A}$ and $ϕ_{B}$ are the work function of A and B respectively

all of the above

answer is C.

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Detailed Solution

From Einstein’s photoelectric equation
$h ϑ = h ϑ_{0} + \frac{1}{2} m ϑ_{\max}^{2} = h ϑ_{0} + e V_{s}$
$V_{s} = (\frac{h}{e}) ϑ - \frac{ϕ}{e}$
So the slope $m = \tan θ = \frac{h}{e}$
the intersection of the line on $y - axis$ is
$\frac{C_{1}}{C_{2}} = \frac{\frac{ϕ_{A}}{e}}{\frac{ϕ_{B}}{e}} = \frac{ϕ_{A}}{ϕ_{B}} (∵ C_{1} < C_{2}) ∴ ϕ_{A} < ϕ_{B}$