Q33

1 markMCQSection D

From the graph, the work functions of A and B are (h is Planck's constant and e value of charge on an electron)

Q: From the graph, the work functions of A and B are (h is Planck's constant and e value of charge on an electron)

**Correct option: (C) $h\nu_1$ and $h\nu_2$** From Einstein's photoelectric equation, $eV_s = h\nu - \phi$. Each line meets the ν-axis $(V_s = 0)$ at the **threshold frequency**: - For metal A the intercept is $\nu_1$, so $\phi_A = h\nu_1$ - For metal B the intercept is $\nu_2$, so $\phi_B = h\nu_2$ The work function equals Planck's constant times the threshold frequency, giving **$h\nu_1$ and $h\nu_2$**.

Q: How do you read the work function off a stopping-potential vs frequency graph?

Use $eV_s = h\nu - \phi$. When the stopping potential $V_s$ is zero, the incident frequency equals the threshold frequency $\nu_0$, and the line crosses the frequency axis there. Since $0 = h\nu_0 - \phi$, the work function is $\phi = h\nu_0$ — Planck's constant times the x-intercept. For metal A that intercept is $\nu_1$ and for B it is $\nu_2$, giving work functions $h\nu_1$ and $h\nu_2$.

Q: Why are the lines for A and B parallel?

The slope of every V_s–ν line is $h/e$, which depends only on Planck's constant and the electron charge — universal constants. So all metals give lines of the same slope; they differ only in where they cross the frequency axis, which reflects their different work functions.

Graph of stopping potential vs frequency for metals A and B

Dual Nature of Radiation and Matter

Work function from V_s–ν graph

Options

(A)

\nu_1

and

\nu_2

(B)

V_1

and

V_2

(C)

h\nu_1

and

h\nu_2

(D)

h\nu_1/e

and

h\nu_2/e

Official Answer

Correct option: (C) $h\nu_1$ and $h\nu_2$

From Einstein's photoelectric equation, $eV_s = h\nu - \phi$ . Each line meets the ν-axis $(V_s = 0)$ at the threshold frequency:

For metal A the intercept is $\nu_1$ , so $\phi_A = h\nu_1$
For metal B the intercept is $\nu_2$ , so $\phi_B = h\nu_2$

The work function equals Planck's constant times the threshold frequency, giving $h\nu_1$ and $h\nu_2$ .

work functionthreshold frequencyphi = h nu_0stopping potential frequency graphEinstein photoelectric equationx-intercept thresholdhν_1 and hν_2

Marking Scheme

11 mark: correct option (C) $h\nu_1$ and $h\nu_2$ .
2Reasoning credit for stating $\phi = h\nu_0$ where $\nu_0$ is the x-intercept (threshold frequency).
3No marks for options with wrong dimensions ( $\nu_1$ , $V_1$ or $h\nu/e$ ).

Hint

Set $V_s = 0$ in $eV_s = h\nu - \phi$ : the x-intercept is the threshold frequency, and $\phi = h \times$ (that intercept).

Quick Oral Answer

Set the stopping potential to zero in $eV_s = h\nu - \phi$ ; the line then crosses the frequency axis at the threshold frequency, so the work function is Planck's constant times that intercept — $h\nu_1$ for A and $h\nu_2$ for B.

Analysis & Explanation

Concept:

Einstein's equation $eV_s = h\nu - \phi$ , written as $V_s = (h/e)\nu - \phi/e$ , is a straight line of slope $h/e$ . Where the line crosses the frequency axis, $V_s = 0$ , so $0 = h\nu_0 - \phi$ , i.e. $\phi = h\nu_0$ . The x-intercept of each line is therefore its threshold frequency.

Reading the graph:

Line A cuts the ν-axis at $\nu_1 \Rightarrow \phi_A = h\nu_1$ .
Line B cuts the ν-axis at $\nu_2 \Rightarrow \phi_B = h\nu_2$ .

Why the other options fail:

(A) $\nu_1, \nu_2$ are frequencies, not energies — dimensionally wrong for a work function.
(B) $V_1, V_2$ are potentials (volts); multiplying by e would give energy, but the plain V-values are not the work function.
(D) $h\nu/e$ has units of volts, again not energy.

Exam trap:

Watch units — a work function is an energy (joule or eV), so only $h\nu_0$ has the right dimensions.

Common Mistakes

1Reading the y-intercept $(-V_1, -V_2)$ as the work function instead of using the x-intercept.
2Quoting $\nu_1$ and $\nu_2$ as the work functions, ignoring that a work function must have units of energy.
3Confusing threshold frequency with the frequency at which the two lines cross.

Interesting Facts

Robert Millikan, who initially doubted Einstein's photon idea, spent a decade measuring these $V_s$ – $\nu$ lines and in 1916 used their slope to give one of the best early values of Planck's constant.

The x-intercept (threshold frequency) is a fixed property of the metal — the two parallel lines never move sideways unless you change the metal.

Caesium, with a work function near 2.1 eV, has such a low threshold that even yellow light ejects electrons, which is why it is used in photocathodes.

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Frequently Asked Questions

How do you read the work function off a stopping-potential vs frequency graph?

Use $eV_s = h\nu - \phi$ . When the stopping potential $V_s$ is zero, the incident frequency equals the threshold frequency $\nu_0$ , and the line crosses the frequency axis there. Since $0 = h\nu_0 - \phi$ , the work function is $\phi = h\nu_0$ — Planck's constant times the x-intercept. For metal A that intercept is $\nu_1$ and for B it is $\nu_2$ , giving work functions $h\nu_1$ and $h\nu_2$ .

Why are the lines for A and B parallel?

The slope of every V_s–ν line is $h/e$ , which depends only on Planck's constant and the electron charge — universal constants. So all metals give lines of the same slope; they differ only in where they cross the frequency axis, which reflects their different work functions.