Verified Solution • 2025–26 Session moderate

QJEEM25Ph1597: JEE Main 2025 Physics - Class 12 JEEM25Ph1597 Solution: In photoelectric effect, the stopping potential frequency curve …

In photoelectric effect, the stopping potential (V0)v/s\left(\mathrm{V}_0\right) \mathrm{v} / \mathrm{s} frequency (v)(v) curve is plotted.

( h is the Planck's constant and ϕ0\phi_0 is work function of metal )

(A) V0v/sv\mathrm{V}_0 \mathrm{v} / \mathrm{s} v is linear.

(B) The slope of V0v/sv\mathrm{V}_0 \mathrm{v} / \mathrm{s} v curve =ϕ0 h=\frac{\phi_0}{\mathrm{~h}}

(C) h constant is related to the slope of V0v/sv\mathrm{V}_0 \mathrm{v} / \mathrm{s} v line.

(D) The value of electric charge of electron is not required to determine h using the V0v/sv\mathrm{V}_0 \mathrm{v} / \mathrm{s} v curve.

(E) The work function can be estimated without knowing the value of hh .

Choose the correct answer from the options given below :

Multiple Choice Options

  • A (A), (C) and (E) only
  • B (C) and (D) only
  • C (A), (B) and (C) only
  • D (D) and (E) only

Step-by-Step Verified Solution

amp;hv=ϕ+KEmax⁡amp;KEmax⁡=eV0amp; V0=hv−ϕe\begin{aligned}& \mathrm{hv}=\phi+\mathrm{KE}_{\max } \\& \mathrm{KE}_{\max }=\mathrm{eV}_0 \\& \mathrm{~V}_0=\frac{\mathrm{hv}-\phi}{\mathrm{e}}\end{aligned}

(A) V0v/sV\mathrm{V}_0 \mathrm{v} / \mathrm{s} \mathrm{V} is linear correct

(B) Slope

v0=(he)v−ϕe Wrong \mathrm{v}_0=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{v}-\frac{\phi}{\mathrm{e}} \text { Wrong }

Slope he\frac{\mathrm{h}}{\mathrm{e}}

(C) Correct

(D) Incorrect

(E) Correct

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