Question 71: Dual Nature of Matter and Radiation

Question

In an experiment with photoelectric effect, the stopping potential,

Accepted Answer

We know, $$e{V_0} = hv - \phi $$ .... (i)
$$ \Rightarrow e{V_0} = K{E_{\max }}$$ (where $\phi$ = work function of the metal and $v_0$ = stopping potential)
$$ \Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}$$
Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.
From eq. (i), we can see,
$$v \uparrow  \Rightarrow {v_0} \uparrow $$
$$ \Rightarrow {e \over {\lambda  \downarrow }} \Rightarrow {v_0} \uparrow $$ (as $$v = {c \over \lambda }$$)
So, stopping potential increases with decrease in the wavelength of the incident light.
Hence, option 1 is correct.

Answer

is $\left(\frac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons

Answer

increases with increase in the intensity of the incident light

Answer

decreases with increase in the intensity of the incident light

Answer

increases with increase in the wavelength of the incident light

Dual Nature of Matter and Radiation

In an experiment with photoelectric effect, the stopping potential,

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