In an experiment with photoelectric effect, the stopping potential. [2025]

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Q.
In an experiment with photoelectric effect, the stopping potential. [2025]

1 increases with increase in the wavelength of the incident light.

2 increases with increase in the intensity of the incident light.

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

4 decreases with increase in the intensity of the incident light.

Ans.
(3)

From Einstein photoelectric equation

$\frac{h c}{λ} = ϕ + {eV}_{S}$

Maximum K.E. $= {(K)}_{\max} = {eV}_{S}$

So, $V_{S} = \frac{{(K)}_{\max}}{e}$

Solution

Q.
In an experiment with photoelectric effect, the stopping potential. [2025]

1 increases with increase in the wavelength of the incident light.

2 increases with increase in the intensity of the incident light.

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

4 decreases with increase in the intensity of the incident light.

Ans.
(3)

From Einstein photoelectric equation

$\frac{h c}{λ} = ϕ + {eV}_{S}$

Maximum K.E. $= {(K)}_{\max} = {eV}_{S}$

So, $V_{S} = \frac{{(K)}_{\max}}{e}$

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Solution

Q. In an experiment with photoelectric effect, the stopping potential. [2025]

1 increases with increase in the wavelength of the incident light.

2 increases with increase in the intensity of the incident light.

3 is (1e) times the maximum kinetic energy of the emitted photoelectrons.

4 decreases with increase in the intensity of the incident light.

Ans. (3) From Einstein photoelectric equation hcλ=ϕ+eVS Maximum K.E. =(K)max=eVS So, VS=(K)maxe

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Q.
In an experiment with photoelectric effect, the stopping potential. [2025]

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

Ans.
(3)

From Einstein photoelectric equation

$\frac{h c}{λ} = ϕ + {eV}_{S}$

Maximum K.E. $= {(K)}_{\max} = {eV}_{S}$

So, $V_{S} = \frac{{(K)}_{\max}}{e}$