The ratio of slopes of K max operatornameVs V and Vo V S ∨ curves in the photoelectric effects gives (v=. frequency, K max = maximum kinetic energy, v0= stopping potential )

Question

The ratio of slopes of K max operatornameVs V and Vo V S ∨ curves in the photoelectric effects gives (v=. frequency, K max = maximum kinetic energy, v0= stopping potential ) (A) the ratio of Planck’s constant of electronic charge (B) work function (C) Planck’s constant (D) charge of electron

Tardigrade Admin · Accepted Answer

h v=h v0+e v0 or, e v0=h v-h v0 or, v0=(h/e) v-(h/e) v0 On comparing the above equation with the straight line equation, i.e y=m x+c The slope of v0 vs v is text (slope )1=(h/e) Similarly, Thus, Slope of Kmax vs v is (slope)2 = h ∴ ((slope)2/(slope)1) = (h/h/e) = e

Q. The ratio of slopes of Kmax​VsV and Vo​VS∨ curves in the photoelectric effects gives (v= frequency, Kmax​= maximum kinetic energy, v0​= stopping potential )

the ratio of Planck’s constant of electronic charge

work function

Planck’s constant

charge of electron

hv=hv0​+ev0​ or, ev0​=hv−hv0​ or, v0​=eh​v−eh​v0​ On comparing the above equation with the straight line equation, i.e y=mx+c The slope of v0​ vs v is (slope )1​=eh​ Similarly, Thus, Slope of KXmax​ vs v is (slope)X2​=h ∴(slope)X1​(slope)X2​​=h/eh​=e

Q. The ratio of slopes of $K_{m a x} Vs V$ and $V_{o} V S \lor$ curves in the photoelectric effects gives $(v =$ frequency, $K_{m a x} =$ maximum kinetic energy, $v_{0} =$ stopping potential $)$