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: