Q36

1 markMCQSection D

The threshold frequency for a metal surface is $\nu_0$ . If the radiation of frequency $3\nu_0$ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is $E_1$ . If the frequency were increased to $6\nu_0$ , the maximum KE of the photoelectrons becomes $E_2$ . Then $(E_1/E_2)$ equals

OR

Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant 'h' is given by

(A) $me$ (B) $1/(me)$ (C) $m/e$ (D) $e/m$

Q: The threshold frequency for a metal surface is $\nu_0$. If the radiation of frequency $3\nu_0$ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is $E_1$. If the frequency were increased to $6\nu_0$, the maximum KE of the photoelectrons becomes $E_2$. Then $(E_1/E_2)$ equals OR Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant 'h' is given by (A) $me$ (B) $1/(me)$ (C) $m/e$ (D) $e/m$

**Correct option: (C) $2/5$** **Primary part — KE ratio:** Using $K_{max} = h(\nu - \nu_0)$: - $E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$ - $E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0$ - **$E_1/E_2 = 2h\nu_0 / 5h\nu_0 = 2/5$** **OR alternative — Planck's constant from slope:** Since $V_s = (h/e)\nu - \varphi/e$, the slope is $m = h/e$, so **$h = me$ → option (A)**.

Q: Why is $E_1/E_2 = 2/5$ and not $1/2$ for frequencies $3\nu_0$ and $6\nu_0$?

Maximum kinetic energy is $K_{max} = h(\nu - \nu_0)$, so you must subtract the threshold frequency before comparing. At 3ν₀, $E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$; at 6ν₀, $E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0$. The ratio is $2h\nu_0/5h\nu_0 = 2/5$. Taking $3/6 = 1/2$ is the classic mistake of forgetting the ν₀ subtraction.

Dual Nature of Radiation and Matter

Ratio of maximum kinetic energies at $3\nu_0$ and $6\nu_0$

Options

(A)

1/3

(B)

1/2

(C)

2/5

(D)

3/4

Official Answer

Correct option: (C) $2/5$

Primary part — KE ratio:

Using $K_{max} = h(\nu - \nu_0)$ :

$E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$
$E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0$
$E_1/E_2 = 2h\nu_0 / 5h\nu_0 = 2/5$

OR alternative — Planck's constant from slope:

Since $V_s = (h/e)\nu - \varphi/e$ , the slope is $m = h/e$ , so $h = me$ → option (A).

K_max = h(ν − ν_0)kinetic energy ratio 2/5threshold frequency subtractionEinstein photoelectric equationslope h/ePlanck constant h = mephotoelectric effect numerical

Marking Scheme

11 mark: correct option (C) $2/5$ for the primary KE-ratio part.
2Full credit for the OR alternative: option (A) $me$ from slope $m = h/e$ .
3No marks for the $1/2$ distractor obtained by forgetting to subtract the threshold frequency.

Hint

Use $K_{max} = h(\nu - \nu_0)$ : subtract the threshold before taking the ratio; for the OR part slope $m = h/e \Rightarrow h = me$ .

Quick Oral Answer

Maximum kinetic energy is h times (ν minus ν-naught), so $E_1 = 2h\nu_0$ at $3\nu_0$ and $E_2 = 5h\nu_0$ at $6\nu_0$ ; their ratio is $2/5$ , and from the graph's slope $h/e$ we get Planck's constant $h = m\cdot e$ .

Analysis & Explanation

Concept:

Einstein's equation in threshold form, $K_{max} = h(\nu - \nu_0)$ , gives the maximum kinetic energy directly as Planck's constant times the excess of frequency over threshold.

Working the primary part:

At $3\nu_0$ : $E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$ .
At $6\nu_0$ : $E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0$ .
Ratio $E_1/E_2 = 2/5$ . The $h\nu_0$ cancels, so the answer is independent of the metal.

OR part:

Rearranging $eV_s = h\nu - \varphi$ gives $V_s = (h/e)\nu - \varphi/e$ , a line of slope $h/e$ .
If that slope is m, then $m = h/e \Rightarrow h = me$ → option (A).

Exam trap:

Students often forget to subtract $\nu_0$ and wrongly take the ratio $3/6 = 1/2$ (distractor B). The threshold subtraction is essential.

Common Mistakes

1Taking the ratio of frequencies $(3\nu_0/6\nu_0 = 1/2)$ without subtracting the threshold frequency $\nu_0$ .
2Forgetting that $K_{max} = h(\nu - \nu_0)$ , not $h\nu$ , so the work function/threshold term is dropped.
3In the OR part, writing $h = m/e$ or $e/m$ instead of $h = me$ from slope $= h/e$ .

Interesting Facts

The relation $K_{max} = h(\nu - \nu_0)$ means a photoelectric plot's slope gives Planck's constant regardless of the metal — Millikan's 1916 measurement this way matched the value from black-body radiation, a triumph for quantum theory.

Because the $h\nu_0$ term cancels in the ratio, $E_1/E_2 = 2/5$ holds for every metal illuminated at $3\nu_0$ and $6\nu_0$ .

Doubling the frequency here ( $3\nu_0 \to 6\nu_0$ ) more than doubles the kinetic energy ( $2h\nu_0 \to 5h\nu_0$ ) precisely because the fixed threshold energy $h\nu_0$ is subtracted first.

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Frequently Asked Questions

Why is $E_1/E_2 = 2/5$ and not $1/2$ for frequencies $3\nu_0$ and $6\nu_0$ ?

Maximum kinetic energy is $K_{max} = h(\nu - \nu_0)$ , so you must subtract the threshold frequency before comparing. At 3ν₀, $E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$ ; at 6ν₀, $E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0$ . The ratio is $2h\nu_0/5h\nu_0 = 2/5$ . Taking $3/6 = 1/2$ is the classic mistake of forgetting the ν₀ subtraction.

How does the OR part give Planck's constant as $h = me$ ?

Rearranging Einstein's equation $eV_s = h\nu - \varphi$ gives $V_s = (h/e)\nu - \varphi/e$ , a straight line with slope h/e. If the measured slope of metal B's line is m, then $m = h/e$ , so Planck's constant $h = me$ — the slope multiplied by the electron charge.