In a photoelectric experiment, the stopping potential Vs is plotted against the frequency v of the incident light. The resulting curve is a straight line which makes an angle θ with the v-axis. Then tanθ will be equal to (Here textE text0 text= work function of the surface)

Question

In a photoelectric experiment, the stopping potential Vs is plotted against the frequency v of the incident light. The resulting curve is a straight line which makes an angle θ with the v-axis. Then tanθ will be equal to (Here textE text0 text= work function of the surface) (A) (h/e) (B) (e/h) (C) -(φ/e) (D) (e h/φ)

Tardigrade Admin · Accepted Answer

EK=hv-φ eVs=hv-φ Vs=(h v/e)-(φ/e) â¦(i) y=mx+C â¦(ii) Comparison of equation (i) and (ii) shows that the slope i.e., tanθ will be equal to (h/e) . m(.Slope.)=tanθ =(h/e) .

Q. In a photoelectric experiment, the stopping potential $V_{s}$ is plotted against the frequency v of the incident light. The resulting curve is a straight line which makes an angle $θ$ with the v-axis. Then $t an θ$ will be equal to

(Here $E_{0} =$ work function of the surface)

$\frac{h}{e}$

$\frac{e}{h}$

$- \frac{ϕ}{e}$

$\frac{e h}{ϕ}$

$E_{K} = h v - ϕ$

$e V_{s} = h v - ϕ$

$V_{s} = \frac{h v}{e} - \frac{ϕ}{e}$ …(i)

$y = m x + C$ …(ii)

Comparison of equation (i) and (ii) shows that the slope i.e., $t an θ$ will be equal to $\frac{h}{e}$ .

$m (Sl o p e) = t an θ = \frac{h}{e}$ .

Q. In a photoelectric experiment, the stopping potential Vs​ is plotted against the frequency v of the incident light. The resulting curve is a straight line which makes an angle θ with the v-axis. Then tanθ will be equal to (Here E0​= work function of the surface)

eh​

he​

−eϕ​

ϕeh​