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Step-by-Step Explanation
1. Understanding the Photoelectric Effect
The photoelectric effect describes how electrons (photoelectrons) are ejected from a metal surface when light of sufficiently high frequency (or sufficiently low wavelength) strikes it. The kinetic energy of these photoelectrons depends on the frequency of the incident light.
2. Relationship Between Wavelength and Frequency
The frequency $ \nu $ of the light is inversely proportional to its wavelength $ \lambda $, given by
$ \nu = \frac{c}{\lambda} $
where $ c $ is the speed of light in vacuum.
3. Kinetic Energy of Photoelectrons
The maximum kinetic energy $K_{\mathrm{max}}$ of the emitted photoelectrons is given by
$ K_{\mathrm{max}} = h \nu - \phi, $
where $ h $ is Planck’s constant and $ \phi $ is the work function of the metal. As $ \nu $ increases (i.e., when $ \lambda $ decreases), $ K_{\mathrm{max}} $ becomes larger.
4. Effect on Photoelectric Current
When the anode voltage is kept fixed but the wavelength $ \lambda $ of the incident radiation is decreased (i.e., frequency $ \nu $ increases), the photoelectrons are emitted with greater kinetic energy. This increases the likelihood of these energetic electrons reaching the anode before being scattered or slowed down.
Consequently, the number of electrons collected at the anode per unit time (the photoelectric current) increases as the wavelength decreases.
5. Conclusion
As the wavelength of the light falling on the cathode is gradually decreased, the speed (and thus the kinetic energy) of the emitted photoelectrons increases. This allows more electrons to be collected at the anode, thereby increasing the photoelectric current. Hence, the correct graph is the one that shows an increase in photoelectric current as $ \lambda $ decreases.
Correct Answer
