In an experiment aimed to disprove Einstein's photoelectric equation,how did Millikan prove it?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Einstein's photoelectric equation is given by:
$\frac{1}{2} m v_{\max }^{2} = h \nu - \phi_{0}$
Since the maximum kinetic energy is related to the stopping potential $V_{0}$ by the relation:
$\frac{1}{2} m v_{\max }^{2} = e V_{0}$
Substituting this into the photoelectric equation,we get:
$e V_{0} = h \nu - \phi_{0}$
Rearranging for $V_{0}$:
$V_{0} = \left( \frac{h}{e} \right) \nu - \frac{\phi_{0}}{e}$
This equation is of the form $y = mx + c$,representing a straight line where the slope is $\frac{h}{e}$.
Millikan performed a series of experiments measuring the stopping potential $V_{0}$ for different frequencies $\nu$ of incident radiation. He plotted a graph of $V_{0}$ versus $\nu$,which resulted in a straight line. The slope of this line was found to be $\frac{h}{e}$.
By using the known value of the elementary charge $e$,Millikan calculated the value of Planck's constant $h$ to be approximately $6.626 \times 10^{-34} \text{ Js}$,which matched the previously accepted value.
Thus,while Millikan initially intended to disprove Einstein's theory,his experimental results provided strong evidence for its validity,confirming the photoelectric equation with great precision across various alkali metals.

Explore More

Similar Questions

The work function of $Zn$ is $4.25 \ eV$. The threshold frequency corresponds to which region of the electromagnetic spectrum?

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).

$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be

The maximum kinetic energy of the photoelectrons varies:

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo