The ratio of slopes of K_{{max}} vs V and V₀ vs v curves in the photoelectric effects gives (v = frequency, K_{{max}} = maximum kinetic energy, v₀ = stopping potential)

Correct Answer is : charge of electron

the ratio of Planck’s constant of electronic charge

Correct Answer is : charge of electron

Test Index

VITEEE 2015 Paper

Section: Chemistry

Question : 68 of 120

Marks: +1, -0

The ratio of slopes of

K_{\text{max}}

V_0

K_{\text{max}}

v_0

= stopping potential)

Solution:

hv =

hv_0+ev_0

v_0

\frac{h}{e} v - \frac{h}{e} v_0

On comparing this equation with the straight line equation, i.e y = mx + c The slope of

v_0

vs v is (

v_0

is stopping potential)

(\text{slope})_1

\frac{h}{e}

Likewise,

hv =

hv_0

K_{\text{max}}

K_{\text{max}}

= hv -

hv_0

Thus, slope of

K_{\text{max}}

vs v is

(\text{slope})_2

= h

∴

\frac{(\text{slope})_1}{(\text{slope})_2}

\frac{h}{\frac{h}{e}}

= e

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