In an experiment with photoelectric effect, the stopping potential,

is (\frac1e) times the maximum kinetic energy of the emitted photoelectrons

increases with increase in the intensity of the incident light

decreases with increase in the intensity of the incident light

increases with increase in the wavelength of the incident light

Dual Nature of Radiation PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q52 with Solution

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Correct answer: A

Relation between stopping potential and maximum kinetic energy

In the photoelectric effect, the maximum kinetic energy of emitted photoelectrons is related to stopping potential $V_0$ by

K_{\max} = eV_0

where $e$ is the electronic charge.

So,

V_0 = \frac{K_{\max}}{e}

This means the stopping potential is indeed $\left(\frac{1}{e}\right)$ times the maximum kinetic energy.

So Option A is correct.

Dependence on intensity of incident light

From Einstein’s photoelectric equation,

K_{\max} = h\nu - \phi

where:

$h\nu$ is the photon energy,
$\phi$ is the work function.

Thus $K_{\max}$ , and hence stopping potential $V_0$ , depends on the frequency of incident light, not on its intensity.

So:

Option B: "increases with increase in intensity" $\to$ False
Option C: "decreases with increase in intensity" $\to$ False

Intensity affects the number of emitted electrons (photoelectric current), not their maximum kinetic energy.

Dependence on wavelength

Since

\nu = \frac{c}{\lambda}

increasing wavelength $\lambda$ decreases frequency $\nu$ , so photon energy $h\nu$ decreases.

Therefore,

K_{\max} = h\nu - \phi

also decreases with increase in wavelength (provided emission still occurs). Hence stopping potential decreases, not increases.

So Option D is false.

Final evaluation of options

A: True
B: False
C: False
D: False

Therefore, the single correct answer is A.

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View written solutionFree

Correct answer: A

Relation between stopping potential and maximum kinetic energy

In the photoelectric effect, the maximum kinetic energy of emitted photoelectrons is related to stopping potential $V_0$ by

K_{\max} = eV_0

where $e$ is the electronic charge.

So,

V_0 = \frac{K_{\max}}{e}

This means the stopping potential is indeed $\left(\frac{1}{e}\right)$ times the maximum kinetic energy.

So Option A is correct.

Dependence on intensity of incident light

From Einstein’s photoelectric equation,

K_{\max} = h\nu - \phi

where:

$h\nu$ is the photon energy,
$\phi$ is the work function.

Thus $K_{\max}$ , and hence stopping potential $V_0$ , depends on the frequency of incident light, not on its intensity.

So:

Option B: "increases with increase in intensity" $\to$ False
Option C: "decreases with increase in intensity" $\to$ False

Intensity affects the number of emitted electrons (photoelectric current), not their maximum kinetic energy.

Dependence on wavelength

Since

\nu = \frac{c}{\lambda}

increasing wavelength $\lambda$ decreases frequency $\nu$ , so photon energy $h\nu$ decreases.

Therefore,

K_{\max} = h\nu - \phi

also decreases with increase in wavelength (provided emission still occurs). Hence stopping potential decreases, not increases.

So Option D is false.

Final evaluation of options

A: True
B: False
C: False
D: False

Therefore, the single correct answer is A.

Dual Nature of Radiation question

Dual Nature of Radiation question

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