Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately $4.1\times10^{-15}$ V s. Given that the charge of an electron is $1.6\times10^{-19}$ C, find the value of the Planck's constant h.

Observations from an experiment on photoelectric effect for the stopping potential by varying the incident fre

Numericals · Q16

4.1\times10^{-15}

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Concept understanding — Einstein's Photoelectric Equation

Building directly on Planck's quantization idea, Einstein in 1905 proposed that light of frequency $\nu$ should itself be thought of as a stream of discrete energy quanta (photons), each carrying energy $h\nu$ . Applying this to the photoelectric effect: when a single photon of energy $h\nu$ strikes a metal surface, it is completely absorbed by one single electron; part of this energy, equal to the metal's work function $\phi_0$ , is spent freeing the electron from the surface, and the remainder becomes the ejected electron's kinetic energy. Conservation of energy for this one-photon-one-electron process gives Einstein's photoelectric equation, $h\nu=\phi_0+\tfrac12mv^2$ ; if the electron loses no energy internally on its way out, it emerges with the maximum possible kinetic energy, giving the most commonly used form $K_{max}=h\nu-\phi_0$ . At the threshold frequency $\nu_0$ , the photoelectron is emitted with essentially zero kinetic energy, so $h\nu_0=\phi_0$ , meaning the threshold frequency is simply the photon frequency whose energy exactly equals the work function.

Plotting $K_{max}$ against $\nu$ gives a straight line whose slope is $h$ and whose $y$ -intercept (extended back to zero frequency) is $-\phi_0$ -- exactly the graph R. A. Millikan produced experimentally for several metals, finding a common slope $h=6.626\times10^{-34}$ Js across every metal tested, giving the first precise experimental measurement of Planck's constant from the photoelectric effect itself. Einstein's equation resolves every one of the photoelectric effect's experimental laws at once: more photons per second (higher intensity) means proportionally more electrons ejected per second, since each absorbed photon liberates exactly one electron; $K_{max}=h\nu-\phi_0$ depends only on frequency, never on intensity, since intensity does not appear in the formula at all; a definite threshold frequency $\nu_0=\phi_0/h$ must exist below which no single photon carries enough energy to free even one electron, however many such photons arrive; and because each photon is absorbed and its energy transferred in one complete, instantaneous event rather than being slowly accumulated, there is no measurable time lag between illumination and emission.

Einstein's photoelectric equation, written for the stopping potential, is $eV_0=h\nu-\phi_0$ , i.e.\n\n $V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$ \n\nThis is a straight line when $V_0$ is plotted against $\nu$ , with SLOPE $=h/e$ (a universal constant, independent of the emitter material) and intercept $=-\phi_0/e$ (material-dependent). Given that the measured slope of this line is $4.1\times10^{-15}$ V s, Planck's constant follows directly:\n\n $h = (\text{slope}) \times e = (4.1\times10^{-15}\text{ V s}) \times (1.6\times10^{-19}\text{ C})$ \n\n $h = 6.56\times10^{-34}\text{ J s}$ \n\n(the units work out to J s since $1\text{ V}\times1\text{ C}=1\text{ J}$ ). This value is close to the accepted precise value of Planck's constant, $6.626\times10^{-34}$ J s, as expected for an experimentally measured slope. This is exactly the kind of measurement -- plotting stopping potential against frequency and reading off the slope -- that historically let Millikan (1909) confirm Einstein's photon hypothesis by extracting a value of h that matched Planck's own black-body value. [!ANSWER] $h \approx 6.56\times10^{-34}$ J s.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$Q. If the frequency of incident light falling on a photosensitive material is doubled, then kinetic energy of the emitted photoelectron will be ______.$
(a) the same as its initial value
(b) two times its initial value
(c) more than two times its initial value
(d) less than two times its initial value
›Reveal solutionSolution

Einstein's photoelectric equation is linear in frequency but has a constant work-function offset.

By Einstein's photoelectric equation, $KE_{max} = h\nu - \phi_0$ . If frequency is doubled to $2\nu$ :

$KE_{max}' = h(2\nu)-\phi_0 = 2h\nu - \phi_0 = 2(h\nu-\phi_0)+\phi_0 = 2\,KE_{max}+\phi_0$

Since the work function $\phi_0 > 0$ , the new kinetic energy $KE_{max}'$ is strictly greater than $2\,KE_{max}$ .

✓Final answer
$more than two times its initial value — option (c).$
CBSE 2022Set ANNUAL1 markMCQ
$Q. If photons of frequency ν are incident on the surfaces of metals A & B of threshold frequencies ν/2 and ν/3 respectively, the ratio of the maximum kinetic energy of electrons emitted from A to that of from B is$
(a) 2 : 3
(b) 3 : 4
(c) √3 : 2
(d) 1 : 1
›Reveal solutionSolution

Using Einstein's equation $K_{max}=h(\nu-\nu_0)$ for each metal and taking the ratio gives $3:4$ .

Max KE of photoelectrons: $K_{max} = h\nu - h\nu_0 = h(\nu-\nu_0)$ .

For metal A, $\nu_{0A}=\nu/2$ :

$K_A = h\left(\nu-\frac{\nu}{2}\right) = \frac{h\nu}{2}$

For metal B, $\nu_{0B}=\nu/3$ :

$K_B = h\left(\nu-\frac{\nu}{3}\right) = \frac{2h\nu}{3}$

Ratio:

$\frac{K_A}{K_B} = \frac{h\nu/2}{2h\nu/3} = \frac{3}{4}$

✓Final answer
$(b) 3 : 4.$

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