Question 17: Dual Nature of Matter and Radiation

Question

$\mathrm{K}_1$ and $\mathrm{K}_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1=3 \lambda_2$, then

Accepted Answer

Let's analyze the problem using the photoelectric effect equation. The maximum kinetic energy of an emitted photoelectron for a given wavelength is given by:
$ K_{	ext{max}} = \frac{hc}{\lambda} - \phi, $
where
Ã¢ÂÂ¢ $h$ is PlanckÃ¢ÂÂs constant,
Ã¢ÂÂ¢ $c$ is the speed of light,
Ã¢ÂÂ¢ $\lambda$ is the wavelength of the incident light, and
Ã¢ÂÂ¢ $\phi$ is the work function of the metal.
For the two cases:
When the wavelength is $\lambda_1 = 3\lambda_2$, the kinetic energy is
$    K_1 = \frac{hc}{\lambda_1} - \phi = \frac{hc}{3\lambda_2} - \phi.    $
For wavelength $\lambda_2$, the kinetic energy is
$    K_2 = \frac{hc}{\lambda_2} - \phi.    $
Notice that the photon energy for wavelength $\lambda_1$ is
$ E_1 = \frac{hc}{\lambda_1} = \frac{hc}{3\lambda_2}, $
and for wavelength $\lambda_2$ it is
$ E_2 = \frac{hc}{\lambda_2}. $
Since $E_1 = \frac{1}{3} E_2,$ we can write:
$ K_1 = \frac{1}{3}\frac{hc}{\lambda_2} - \phi. $
Now, letÃ¢ÂÂs compare $K_1$ with $\frac{1}{3} K_2$. First, calculate $\frac{1}{3}K_2$:
$ \frac{1}{3}K_2 = \frac{1}{3} \left(\frac{hc}{\lambda_2} - \phiight) = \frac{hc}{3\lambda_2} - \frac{\phi}{3}. $
Subtracting this from the expression for $K_1$ gives:
$ K_1 - \frac{1}{3}K_2 = \left(\frac{hc}{3\lambda_2} - \phiight) - \left(\frac{hc}{3\lambda_2} - \frac{\phi}{3}ight) = -\phi + \frac{\phi}{3} = -\frac{2\phi}{3}. $
Since the work function $\phi$ is positive, it follows that
$ K_1 - \frac{1}{3}K_2 = -\frac{2\phi}{3} < 0. $
Thus, we have:
$ K_1 < \frac{1}{3}K_2. $
This corresponds to Option D.

Answer

$
\mathrm{K}_1=\left(\frac{1}{3}\right) \mathrm{K}_2
$

Answer

$
\mathrm{K}_1>3 \mathrm{~K}_2
$

Answer

$
\mathrm{K}_1>\left(\frac{1}{3}\right) \mathrm{K}_2
$

Answer

$
\mathrm{K}_1<\left(\frac{1}{3}\right) \mathrm{K}_2
$

Dual Nature of Matter and Radiation

\mathrm{K}_1$ and $\mathrm{K}_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1=3 \lambda_2$, then

Choose the correct answer: